vedanta Excel In Opt. Mathematics - Book 9 Trigonometric Ratios
Now, sinT = k,
cosT = b = 1 – k2 = 1 – sin2T
h 1
tanT = p =k k2 = sinT
b =11k– = 1 – sin2T
1
cosecT = h sinT
p
secT = h = 1= 1
b 1 – k2 1 – sin2T
cotT = b = 1 – k2 = 1 – sin2T
p 1 sinT
Alternate method :
By basic trigonometric ratios,
Trigonometric ratios of angle T in term of sinT
cosT = 1 – sin2T
cosT = sinT = sinT
cosT 1 – sin2T
1
cosecT = sinT
secT = 1 = 1
cosT 1 – sin2T
cotT = cosT = 1 – sin2T
sinT sinT
Example 2. If cosT = 4 , find the value of other remaining trigonometric ratios.
Solution: 5
4 b
Example 3. Here, cosT = 5 = h
Solution:
If b = 4, h = 5, then p = h2 – b2 = 25 – 16 = 9 = 3
Now, sinT = p = 3
h 5
p
tanT = b = 3
4
hp 5
cosecT = = 3
secT = h = 5
b 4
b 4
cotT = p = 3
If tanD = 1, find the value of 3sinD – 4sin2D.
Here, tanD = 1 = p
b
If p = 1, b = 1,
then, h = p2 + b2 = 1 + 1 = 2
sinD p = 1
h 2
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vedanta Excel In Opt. Mathematics - Book 9 Trigonometric Ratios
( )Now, 3 sinD – 4sin3D 13
= 3. 1 – 4 .
2 2
3 – 4 6–4 2 1
= 2 22 = 22 2 2 = 2
Example 4. If (a + 1) sinD = (a – 1), show that 2 a . tanD = a – 1
Solution: Here, (a + 1) sinD = (a – 1)
i.e. sinD = a–1 = p
a+1 h
If p = a – 1, h = a+ 1,
then, b = h2 – b2
= (a + 1)2 – (a – 1)2
= a2 + 2a + 1 – a2 + 2a – 1 =2 a
Now, tanD = p .
b
a–1
or, tanD = 2a
? 2 a tanD = a – 1 proved.
Example 5. If sinT + sin2T = 1, show that cos2T + cos4T = 1.
Solution : Here, sinT + sin2T = 1
or, sinT = 1 – sin2T
or, sinT = cos2T
Square on both sides, we get
sin2T = (cos2T)2
or, 1 – cos2T = cos4T
or, cos4T + cos2 = 1
? cos4T + cos2T = 1 proved.
Example 6. If 5cosT + 12sinT = 13, prove that tanT =13 .
Solution:
We have, 5 cosT + 12 sinT = 13
Dividing both sides by cosT, we get
5 cosT + 12 sinT = 13
cosT cosT cosT
or, 5 + 12 tanT = 13secT
Squaring on both sides, we get,
(5 + 12tanT)2 = (13secT)2
or, 25 + 120 tanT + 144 tan2T = 169 sec2T
or, 144 tan2T + 120 tanT + 25 – 169 (1 + tan2T) = 0
or, – 25 tan2T + 120 tanT – 144 = 0
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vedanta Excel In Opt. Mathematics - Book 9 Trigonometric Ratios
or, (5tanT)2 – 2.5. tanT. 12 + (12)2 = 0
or, (5tanT – 12 )2 = 0
? tanT = 12 proved.
5
Example 7. If sinx + cosx = 2 cosx, prove that cosx – sinx = 2 . sinx.
Solution:
Here, sinx + cosx = 2 cosx
Example 8.
Solution: Squaring on both sides, we get,
(sinx + cosx)2 = 2cos2x
or, sin2x + 2sinx.cosx + cos2x = 2cos2x
or, 1 – cos2x + 2sinx .cosx + 1 – sin2x = 2 – 2sin2x
or, – cos2x + 2sinx.cosx – sin2x = – 2sin2x
or, cos2x – 2sinx .cosx + sin2x = 2sin2x
or, (cosx – sinx)2 = ( 2 sinx)2
? cosx – sinx = 2 sinx proved.
Alternative Method
Here, sinx + cosx = 2 cosx
or, sinx = 2 cosx – cosx
or, sinx = ( 2 – 1) cosx
or, sinx = 2– 1 × 2 + 1. cosx
2+ 1
2–1
or, sinx = 2+1 cosx
or, 2 sinx + sinx = cosx
or, cosx – sinx = 2 sinx
? cosx – sinx = 2 sinx proved.
If sinD = s35inaDn=d si35nE== 5 , find the value of tanD + tanE
We have, p13 1 – tanD.tanE
h
If p = 3, h = 5, b = h2 – p2 = 25 – 9 = 16 = 4
? tanD = p = 3
b 4
5 p
Again, sinE = 13 = h
If p = 5, h = 13, b = h2 – p2 = 169 – 25 = 144 =12
tanE = p = 5
b 12
3 5
tanD + tanE 4 + 12
1 – tanD.tanE
Now, = 1– 3 . 5
4 12
= 9+5 × 16 5 = 14 × 16 = 56
12 16 – 12 11 33
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vedanta Excel In Opt. Mathematics - Book 9 Trigonometric Ratios
Exercise 9.3
Very short Questions
1. (a) If sinT = 3 , find the value of tanT.
2
(b) If tanT = 1, find the value of sinT.
(c) If cosT = 1 , find the value of sinT.
2
(d) If secT = 2 , find the value of sinT.
3
2x
2. (a) If sinT = x+ 1 , find the value of tanT.
(b) If cosT = x x 1 , find the value of cotT.
+
Short Questions
3. (a) If tanT = 3 , find the values of cosT and cosecT.
4
(b) If 3tanT = 2, find the value of sinT and cotT.
(c) If sinT = cosT, find the values of sinT, cosT and tanT.
(d) If 5 tanT = 4, find the value of 5sinT – 3cosT
5sinT + 2cosT
(e) If tanT = x , find the value of xsinT – ycosT
y xsinT + ycosT
(f) If sinT = a–1 , find the value of tanT.
a+1
(g) If sinT + cosecT = 3, find the value of sin2T + cosec2T
4. (a) If sinT = p , show that q2 – p2 tanT = p
q
(b) If cotT = p , show that p2 + q2 cosT = p
q
(c) If cosT = x , prove that xsinT = ycosT.
x2 + y2
tanT 360
(d) If 41 sinT = 40, show that tan2T – 1 = 1519
(e) If 1 – cosT = 1 , show that tan2T + 4sin2T = 6
2
5. Express all trigonometric ratios of an angle T in terms of-
(a) cosT (b) tanT
(c) secT (d) cosecT
(e) cotT
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Long Questions
6. (a) If 5 cosT + 12sinT = 13, find the value of tanT
(b) If 8sinT = 4 + cosT, find the value of tanT
7. (a) If 3 sinT + 4cosT = 5, prove that tanT = 3
4
(b) If 4cos2T + 4sinT = 5, show that sinT = 1
2
(c) If tanT + secT =x, show that sinT = x2 – 1
x2 + 1
(d) If sinT – cosT = 0, show that cosecT = ± 2
(e) If sinT + cosT = 2 cosT, prove that cosT – sinT = 2 sinT.
(f) If tanT + sinT = x, and tanT – sinT = y, show that x2 – y2 = 4 xy
(g) If 1 + sin2T = 3sinT.cosT, prove that tanT = 1 or 1 .
2
8. If cos4T + cos2T = 1, show that
(a) tan4T + tan2T = 1 (b) cot4T – cot2T = 1
9. If tanT = 2xy , prove that sinT = 2xy
x2 – y2 x2 +y2
p pcosT – qsinT p2 – q2
10. If cotT = q , prove that pcosT + qsinT = p2 + q2
1. (a) 3 (b) 1 (c) 3 (d) 1
2 2 2
2. (a) 2x (b) x
x–1 2x + 1
3. (a) 4 , 5 (b) 2 , 3 (c) 1 , 1 , 1 (d) 1
5 3 13 2 2 2 6
(e) x2 – y2 (f) a–1 (g) 7
x2 + y2 2a
5. Show to your teacher
6. (a) 12 (b) 3 or – 5
5 4 12
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vedanta Excel In Opt. Mathematics - Book 9 Trigonometric Ratios of Some Standard Angles
Trigonometric Ratios of
10Some Standard Angles
10.0 Introduction
Group discuss the following questions :
(a) What is the standard position of a triangle ?
(b) Define a unit circle.
(c) Under which condition does a revolving line make 0° and 90° with X-axis?
(d) What is the measure of each angle of an equilateral triangle?
(e) If an angle bisector of vertical angle in an equilateral triangle is drawn, is it a
perpendicular bisector of the base ?
A triangle in which one of the vertices is at the origin and one side is along X-axis is known
as standard position of a triangle. If a circle is drawn with centre at the origin and radius of
1 unit, then the circle is called unit circle.
10.1 Trigonometric ratios of standard Angles
Here, we discuss the trigonometric ratios of angles 0°, 30°, 45°, 60°, and 90°.
(a) Trigonometric ratios of 0° and 90° Y
In the figure, a revolving line OP makes an angle of P
XOP with an intial line OX. PM perpendicular is M
drawn to OX. A circle is drawn with radius OP.
If the revolving line OP overlaps with OX , X' T X
XOP = 0°. If OP = OM then PM = 0. The trigonometric O
ratios of 0° are given below :
Y'
sin0° = PM = 0 =0
OP OP
OP
cos0° = OM = OP =1
OP
PM 0
tan0° = OM = OM = 0
cot0° = 1 = 1 = ∞ (undefined)
tan0° 0
1 1
sec0° = cos0° = 1 =1
cosec0° = 1 = 1 =∞ (undefined)
sin0° 0
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Again if the revolving line OP overlaps with Y-axis, XOP = 90°
i.e. XOY = XOP = 90°, then OP = PM, and OM = 0 .
Now, in right angled triangle POM,
sin90° = PM = OP =1
OP OP
OM 0
cos90° = OP = OP =0
tan90° = PM = OP =∞
OM 0
OM 0
cot90° = PM = OP =0
sec90° = 1 = 1 =∞
cos90° 0
1 1
cosec90° = sin90° = 1 = 1
(b) Trigonometric Ratios of 30° and 60°
In the figure, a revolving line OP makes an angle of Y
30° with the initial line OX i.e. XOP = 30°. A circle
is drawn with radius OP. PM is drawn perpendicular P
to OX. PM is produced to Q. Then, by RHS axiom
of congruency 'POM and 'QOM will be congruent. X' 60° X
Then XOP = XOQ = 30° and ∆ POQ is an 30° M
equilateral triangle.
O
Then, OPM = OQM = POQ = 60° Q
Y'
Let OP = OQ = PQ = 2a
PM = 1 PQ = 1 × 2a = a
2 2
By using Pythagoras theorem, from ∆POM,
OM = OP2 – PM2 = 4a2 – a2 = 3 .a
From right angled triangle POM, taking trigonometric ratios of 30°, we get,
sin30° = PM = a = 1
OP 2a 2
cos30° = OM = 3.a = 3
OP 2a 2
tan30° = PM = a = 1
OM 3a 3
cosec30° = 1 = 1 =2
sin30° 1
sec30° = 1 = 1 2 = 2
cos30° 3 3
1 2 1
tan30° 1
cot 30° = = =3
3
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vedanta Excel In Opt. Mathematics - Book 9 Trigonometric Ratios of Some Standard Angles
Again from right angled triangle POM, taking OPM = 60° as a reference angle, we get
the trigonometric ratios of 60°.
sin60° = OM = 3a = 3
OP 2a 2
cos60° = PM = a = 1
OP 2a 2
tan60° = OM = 3a = 3
PM a
cosec60° = 1 = 1 = 2
sin60° 3 3
sec60° = 1 = 1 2= 2
cos60° 1
cot60° = 1 = 22
tan60° 3
(c) Trigonometric Ratios of 45° Y
P
In the figure, a revolving line OP makes an angle of
45° with the initial line OX. Draw a circle with radius 45° M X
OP. Draw PM ꓕ OX. Then ∆POM is an isoceles right
angled triangle them OM = PM. O
X'
Let OM = PM = a. Then, by using Pythagoras
theorem, we get,
OP = OM2 + PM2 Y'
= a2 + a2
= 2a2 = 2 . a
Now, from right angled triangle POM, we take the trigonometric ratios of 45°.
sin45° = PM = a = 1
OP 2a 2
cos45° = OM = a = 1
OP 2 .a 2
tan45° = PM = a =1
OM a
cosec45° = 1 = 2
sin45°
sec45° = 1 = 2
cos45°
cot45° = 1 =1
tan45°
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vedanta Excel In Opt. Mathematics - Book 9 Trigonometric Ratios of Some Standard Angles
Table
Trigonometric ratios of standard angles
Ratio Angle 0° 30° 45° 60° 90°
1
sin 0 1 2 3 1
2 1 2
2
cos 1 3 1 1 0
tan 0 2 2
1 2
3 3∞
2
cosec ∞ 2 2 1
sec 1 2 1 3
3
2∞
cot ∞ 3 1 0
3
Worked out Examples
Examples 1. Evaluate : (sin60° + cos30°) . cot60°
Solution:
( )Here, (sin60° + cos30°) cot60° = 3 + 3 1
Example 2. 2 2 .3
Solution:
= 3+ 3 = 2 3 =1
Example 3. 23 2 3
Solution:
Evaaluate : cos230° + sin260° – tan260°
( ) ( )Here, cos230° + sin260° – tan260° = 3 2+ 3 2 – ( 3 )2
2 2
3 3
= 4 + 4 – 3
= 3 + 3– 12 = 6 – 12 = – 6 = – 3
4 4 4 2
Evaluate : sec2 Sc sec2 Sc (cosec Sc – cosec Sc )
4 3 6 2
Here, Sc = 180°
( )sec2 Sc Sc Sc Sc
4 sec2 3 cosec 6 – cosec 2
= sec2 180° . sec2 180° (cosec 180° – cosec 180° )
4 3 6 2
= sec245°.sec260°(cosec30° – cosec90°)
= ( 2 )2 (2)2 (2 – 1)
= 2.4.1 = 8
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vedanta Excel In Opt. Mathematics - Book 9 Trigonometric Ratios of Some Standard Angles
Example 4. If T = 30°, verify that sin3T = 3sinT – 4 sin3T .
Solution:
Here, T = 30°
LHS = sin3T = sin(3 . 30°) = sin90° = 1
RHS = 3 sinT – 4sin3T
= 3. sin30° – 4 (sin30°)3
( )= 3. 1 –4 13
2 2
= 3 – 1 = 3–1
2 2 2
2
= 2 =1
? LHS = RHS proved.
Example 5. ( ) ( )Prove that Sc + Sc Sc – Sc = cos60°
Solution: sin 6 cos 6 sin 3 cos 3
Example 6. Here, Sc = 180°
Solution:
( ) ( )Now, LHS = Sc Sc Sc Sc
sin 6 + cos 6 sin 3 – cos 3
( ) ( )=
sin 180° + cos 180° sin 180° – cos 180°
6 6 3 3
= (sin30° + cos30°) (sin60° – cos60°)
( ) ( )= 1 + 3 3 – 1
2 2 2 2
( ) ( )=
1+ 3 3–1
2 2
= ( 3 + 1)( 3 – 1)
4
3 – 1 2 1
= 4 = 4 = 2
RHS = cos60° = 1
2
? LHS = RHS proved.
If D = 60°. E = 30°, verify that tan(D – E) = tanD – tanE
1 + tanD.tanE
Here, D = 60°, E = 30°
Now, LHS = tan (D – E)
= tan(60° – 30°) = tan30° = 1
3
tanD – tanE tan60° – tan30°
RHS = 1 + tanD. tanE = 1 + tan60°.tan30°
3 – 1 3 –1
+ 3 3 =3
= = 2 × 1 = 1
1 1+1 3 2 3
1 × 3
? LHS = RHS proved.
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Example 7. Solve the equation for x : xsin60°. tan60° + sin45°.cos45° + sin230° = sec260°
Solution:
Here, x sin60°.tan60° + sin45°.cos45° + sin230° = sec260°.
or, x. 3 . 3+ 1 . 1 + 1 2 = 22
2 2 2 2
3x 1 1
or, 2 + 2 + 4 =4
or, 3x + 2 + 1 =4
2 4
3x 3
or, 2 + 4 =4
or, 3x = 4 – 3
2 4
3x 16 – 3
or, 2 = 4
or, 3x = 13
2 4
13 × 2
or, 3x = 4
? x = 13
6
Exercise 10
Very short Questions
1. Find the values of the following:
(a) (i) tan60°.sec30° (ii) tan30°.cot60°
(iii) sec60°.cosec45°
(b) Evaluate the following :
(i) san0° + cos60° + sin30° (ii) sin260° + cos260° – cos230°
(iii) cos245° – sec245° + tan260°
(c) Evaluate the following :
(i) sin2 Sc + cos2 Sc (ii) cos Sc + sin Sc + tan Sc
3 6 2 2 4
Sc Sc Sc Sc Sc
(iii) 4 sin2 4 + 3 cos2 3 (iv) tan2 3 + cos2 4 + tan2 4
Short Questions
2. Find the values of the following :
(a) 2sin 60°. sin45° + cos0°. cos45°
(b) 2sin230° – 3cos245° + tan2 30°
(c) 3tan245° – sin260° + 1 . tan2 60° + 2.sin245°
3
(d) 4sin260° + 3tan260° – 16. sin30°.sin45°.cos45°
(e) cos60° – cos245°
sec30°
(f) sin260° + sin230°
1 – 4sin60°.tan60°
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vedanta Excel In Opt. Mathematics - Book 9 Trigonometric Ratios of Some Standard Angles
(g) sin245° – sin260°
cos30° + cos45°
(h) cos60° + sin30°
cot45°
(i) tan45° + sin30°
tan60° + tan30°
3. Evaluate the following :
(a) tan2 Sc + cos2 Sc + sin2 S – cosec2 S
4 4 3 6
(b) sin2 Sc + sin2 Sc + cos2 Sc + cos2 S
4 6 4 3
(c) tan Sc – tan Sc
3 6
1+ tan Sc . tan Sc
3 6
Sc Sc
tan 3 + sin 3
(d) tan Sc . cos Sc
3 4
Sc Sc Sc Sc
(e) cos 3 . cos 4 + sin 4 . sin 3
4. Prove the following :
(a) 2tan30° = 2sin30°.cos30°
1 + tan230°
1 – tan230°
(b) cos60° = 1 + tan230°
(c) sin260° – cos260° = 1
2
2tan60°
(d) 1 – tan260° =– 3
(e) 1 + tan30° = 1 + sin60°
1 – tan30° 1 – sin30°
(f) (1 – sin230°) (1 + cos260°) = 15
(3 – tan245°) (3 + cot245°) 128
(g) 4sin30°.cos60°.sin90°.tan45° = 1.
(h) tan230° + 2sin60° + tan45° – tan60° + cos230° = 25
12
tan260° – cos260° 11
(i) tan45°.sin260° = 3
5. If D = 45°, verify the following :
(a) sin2D = 2sinD.cosD
(b) cos2D = 1 – 2sin2D = cos2D – sin2D
(c) cos2D = 1 – tan2D
1 + tan2D
2tanD
(d) tan2D = 1 – tan2D
(e) sin2D = 1 2tanD
+ tan2D
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6. If D = 60° and E = 30°, verify the following :
(a) sin(D + E) = sinD.cosE + cosD.sinE
(b) sin (D – E) = sinD.cosE – cosD.sinE
(c) tan (D – E) = tanD – tanE
1 + tanD . tanE
(d) cos (D + E) = cosD.cosE – sinD.sinE
7. Solve the following equation for x.
(a) tan245° – cos260° = x sin45°.cos45°.tan60°
(b) x sin45°.tan60°.cos45° = tan245° – cos260°
(c) sin230° + xcot245° = 1 + cos245°
4
(d) sin30° + 2cot230° + xcos230° = 8 + tan245° + cos60°.
(e) x 3 cot30° + tan60° . cosec60° = sin30° + 2cot230° + xcos230°
(f) 4sin260° + 2sec245° = 4cos230° + 3tan230° + x cot230°
(g) tan60°.cosec60° + 3x cot30° = sec60°. cosec30°
1. (a) (i) 2 (ii) 1 (iii) 2 2
3
1 3
(b) (i) 1 (ii) 4 (iii) 2
(c) (i) 3 (ii) 2 (iii) 11 (iv) 9
2 4 2
3+1 2 17
2. (a) 2 (b) – 3 (c) 4 (d) 8
(e) 0 (f) – 1 (g) 1 ( 2– 3) (h) 1 (i) 33
5 2 8
7 3 1 32 3+1
3. (a) – 4 (b) 2 (c) 3 (d) 2 (e) 22
7. (a) 3 (b) 3 (c) 1 (d) 4
2 2 2
2
(e) 2 (f) 1 (g) 3
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vedanta Excel In Opt. Mathematics - Book 9 Trigonometric Ratios of Any Angle
Trigonometric Ratios of
11Any Angle
11.0 Review
Group discuss the following questions
(a) What are complementary and supplementary angles ? Clarify them with examples.
(b) In which quadrants do the angles 60°, 135°, 225°, and Y First Quadrant
Second Quadrant P1
315° lie? Draw diagrams to show them in quadrants.
P2
(c) In the adjoining figure, how are angles made by the
revolving lines OP1, OP2, OP3, and OP4 measured as X' O X
positive angles ?
(d) In which quadrant does the angle 420° lie ?
(e) What is the difference between –30° and +30° ? P3 P4
(f) If we write 780° = 2 × 360° + 60°, in which quadrant Third Quadrant Fourth Quadrant
does the angle 780° lie ? Y'
Trigonometric Ratio in terms of radius of a circle. Y
Let XX' and YY' be two mutually perpendicular lines which P(x, y)
intersect at O called origin. Then, XOX' and YOY' are called
X-axis and Y-axis respectively. ry X
T
Let P(x, y) be any point on the circumference of the circle with O xM
radius r and centre at the origin. Draw PM perpendicular to X'
OX, and let XOP = T.
Now, we have,
Reference angle = XOP = T Y'
Hypotenuse = radius = r = (h)
Perpendicular = PM = y = (p)
Base = OM = x = (b)
Let us take six fundamental trigonometric ratios in right angled ∆ POM,
sinT = p = PM = y i.e. sinT = y-coordinate of P = y
h OP r radius r
b OM x x-coordinate of P x
cosT = h = OP = r i.e. cosT = radius = r
tanT = p = PM = y i.e. tanT = y-coordinate of P = y
b OM x x-coordinate of P x
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cosecT = h = r i.e. cosecT = radius of P = r
p y y-coordinate y
h r radius r
secT = b = x i.e. secT = x-coordinate of P = x
cotT = b = x i.e. cotT = x-coordinate of P = x
p y y-coordinate of P y
11.1.1 Trigonometric Ratios of Negative angle
Let (a, b) be any point on the circle with radius OM = r and centre O. Let XOP = T
Now, sinT = y-coordinate of P = b Y
radius r P(a, b)
x-coordinate of P a
cosT = radius = r
tanT = y-coordinate of P = b T
x-coordinate of P a -T
X' O M X
Let P(a,b) be reflected on X-axis,
P(a, b) X-axis P'(a, –b) P'(a, -b)
Then MOP' = – T
Now, sin(– T) = y-coordinate of P' Y'
cos(– T) radius
tan(–T) b b
=– r = – r = – sinT
= x-coordinate of P'
radius
a
= r = cosT
= y-coordinate of P' = –b = – tanT
x-coordinate of P' a
Similarly, we can show that -
cosec(–T) = – cosecT
sec(–T) = secT
cot (–T) = –cotT.
Hence, the trigonometric ratios of negative angles are tabulate below.
sin(–T) = –sinT, cos(–T) = cosT
tan(–T) = –tanT cosec(– T) = – cosecT
sec(–T) = secT cot(–T) = – cotT
11.1.2 Complementary Angles
Two angles are said to be complementary angles if their sum is 90°. Example 60° and 30°
complementary angles as 60° + 30° = 90°.
Similarly 50° and 40° are complementary angles. (90° – T) and T are also complementary
angles.
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vedanta Excel In Opt. Mathematics - Book 9 Trigonometric Ratios of Any Angle
Trigonometric Ratios of Complementary Angles
To find trigonometric ratios of (90° – T)
Let O be the origin and draw a circle with radius OM = r take P(a, b) point on the circle such
that MOP = T Y
P'(b, a)
Now, sinT = y-coordinate of P = b
radius r
cosT = x-coordinate of P = ar T P(a, b)
radius
y-coordinate of P b
tanT = x-coordinates of P = a X' O T M X
cotT = x-coordinate of P = a
y – coordinate of P b
Reflect P(a, b) on line y = x, we get
P(a, b) o P(b, a) Y'
MOP' = 90° – T
Now, sin (90° – T) = y-coordinate of P' = ar = cosT
radius
x-coordinate of P'
cos (90° – T) = radius = b = sinT
r
tan (90° – T) = y-coordinate of P' = a = cotT
x-coordinate of P' b
Similarly, we can show cosec (90° – T) = secT, sec (90° – T) = cosecT, cot (90° – T) = tanT.
Hence, the trigonometric ratios of complementary angles are tabulated below :
sin (90° – T) = cosT cos (90° – T) = sinT
tan (90° – T) = cotT cot (90° – T) = tanT
cosec (90° – T) = secT sec (90° – T) = cosecT
Alternative Method
Trigonometric ratios of complementary angles:
Let O be the centre of the circle with radius OP = r, where P(a, b) is any point on the circle
and let MOP = T, then MPO = 90° – T.
Now, sinT = PM = b Y
OP r P(a, b)
cosT = OM = a
OP r
tanT = PM = b 90°-T
OM a
X' T X
PM b OM
Again, cos(90° – T) = OP = r = sinT
sin (90° – T) = OM = a = cosT
OP r
cot (90° – T) = PM = b = tanT Y'
OM a
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Similarly we can show that
cosec (90° – T) = secT
tan (90° – T) = cotT
Hence, the trigonometric ratios of complementary angles are tabulated below :
sin(90° – T) = cosT, cos(90° –T) = sinT
tan(90° – T) = cotT cot(90°– T) = tanT
cosec(90° – T) = secT sec(90° – T) = cosecT
(c) Trigonometric ratios of supplementary angles
Two angles are said to be supplementary if their sum is 180°. Example 150° and 30°,
180° – T and T are supplementary angles. Here we find the trigonometric ratios of
supplementary angles.
Trigonometric Ratios of Supplementary Angles
To find trigonometric ratios of (180° – T)
Let us take any point P(a, b) on the circumference of the circle whose radius is OM = r and
centre at the origin. Let MOP = T Y
sinT = y-coordinate of P = b
radius r
cosT = x-coordinate of P = a P'(-a, b) P(a, b)
radius r X' MX
tanT = y-coordinate of P = b TT
x-coordinate of P a O
Let P(a, b) be reflected on Y-axis.
Then P'OX' = T and MOP' = 180° – T.
P(a, b) o P' (–a, b) Y'
sin (180° – T) = y-coordinate of P' = b = sinT
radius r
cos (180° – T) = x-coordinate of P' = –a = – ar = – cosT
radius r
tan (180° – T) = y-coordinate of P' = b = – b =– tanT
radius –a a
Similarly, we can show that-
cosec (180° – T) = cosecT
sec(180° – T) = – secT
cot (180° – T) = – cotT
Hence, we can tabulate above results in a table :
sin(180° – T) = sinT, cos(180° –T) = – cosT
tan(180° – T) = – tanT cot(180°– T) = – cotT
cosec(180° – T) = cosecT sec(180° – T) = –secT
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(d) Trigonometric Ratios of (90° + T)
Let O be the origin and taking O as a centre and OM = r radius, a circle is drawn. Let
P(a, b) be any point on the circle. Y
Let MOP = T P'(–b, a) P(a, b)
Then,
sinT = y-coordinate of P = b X' T MX
radius r O
cosT = x-coordinate of P = a
radius r
tanT = y-coordinate of P = b
x-coordinate of P a
Let P(a, b) be rotated through +90° about the origin. Y'
Then MOP' = 90° + T
P(a, b) o P' (–b,a)
Then,
sin (90° + T) = y-coordinate of P' = a = cosT
radius r
cos (90° + T) = x-coordinate of P' = –b = –sinT
radius r
tan (90° + T) = y-coordinate of P' = a = – cotT
x-coordinate of P' –b
Similarly, we can show that-
cosec (90° + T) = secT
sec (90° + T) = – cosecT
cot (90° + T) = –tanT
Hence, the above results can be tabulated below :
sin (90° + T) = cosT cos (90°+ T) = – sinT
tan (90° + T) = – cotT cot (90° + T) = – tanT
cosec (90° + T) = secT sec (90° + T) = – cosecT
(e) Trigonometric Ratios of (180° + T)
Let O be the origin. A circle is drawn about the centre O and taking radius OM = r, on
X-axis. Take P(a,b) be any point on the circle such that MOP = T Y
Then, P(a, b)
O T MX
sinT = y-coordinate of P = b
radius r Y'
cosT = x-coordinate of P = a X'
radius r P'(-a, -b)
tanT = y-coordinate of P = b
x-coordinate of P a
Let P(a. b) be rotated through 180° about the origin in
positive direction. Then,
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P(a, b) o P' (–a, – b)
and MOP' = 180° + T
sin (180° + T) = y-coordinate of P' = –b = – sinT
radius r
cos (180° + T) = x-coordinate of P' = –a = – cosT
radius r
tan (180° + T) = y-coordinate of P' = –b = b = tanT
x-coordinate of P' –a a
similarly we can show that
cosec (180° + T) = – cosecT
sec (180° + T) = – secT
cot (180° + T) = cotT.
Hence, we can tabulated above results in a table :
sin(180° + T) = – sinT cos(180° + T) = –cosT
tan(180° + T) = tanT cot(180°+ T) = cotT
cosec(180° + T) = – cosecT sec(180° + T) = – secT
(f) Trigonometric Ratios of (270° – T)
Let P(a,b) be any point on the circle whose centre is at the origin and radius OM = r
such that MOP = T Y
sinT = y-coordinate of P = b
cosT = x-coorraddiniuaste of P r
tanT = y-coorrdadiniuatse of P a P(a, b)
x-coordinate of P = r 270°– T
= b X' T MX
a O
Let the point P(a, b) be reflected on y = –x, then T
P(a, b) o P'(–b, – a) P'(-b, -a)
Then, P'OY' = T and MOP' = 270° – T Y'
Now,
sin (270° – T) = y-coordinate of P' = –a = – cosT
radius r
cos(270° – T) = x-coordinate of P' = –b = – sinT
radius r
tan (270° – T) = y-coordinate of P' = – a = cotT
x-coordinate of P' – b
Similarly, we can show that
cosec (270° – T) = – secT
sec (270° – T) = – cosecT
cot (270° – T) = tanT
Hence the above relations can be tabulated as below :
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sin (270° – T) = – cosT cos (270°– T) = – sinT
tan (270° – T) = cotT cot (270° – T) = tanT
cosec (270° – T) = –secT sec (270° – T) = – cosecT
(g) Trigonometric ratios (270° + T)
Let P(a,b) be any point on the circle with centre O and radius OM = r such that MOP = T
Then, sinT = b , cosT = a , tanT = b
r r a
Let the point P(a, b) be rotated through + 270° about the origin.
P(a, b) o P'(b, – a) Y
Then, MOP' = 270° + T.
sin (270° + T) = y-coordinate of P' = –a = – cosT X' T P(a, b)
cos (270° + T) = x-coorraddiniuaste = r = sinT O MX
tan (270° + T) = y-coorrdaidniuatse of P' = b = – cotT
x-coordinate r T
of P'
of P' – a
b
Similarly, we can show that- P'(b, -a)
Y'
cosec (270° + T) = – secT
sec (270° + T) = cosecT
cot (270° + T) = – tanT.
Hence, we can tabulated above result in a table :
sin(270° + T) = – cosT cos(270° + T) = sinT
tan(270° + T) =–cotT cot(270° + T) = – tanT
cosec(270° + T) = –secT sec(270° + T) = cosecT
(h) Trigonometric Ratios of (360° – T)
Let P(a,b) be any point on the circle with radius OM = r and centre at the origin such
that MOP = T Y
Then, sinT = b , cosT = a , tanT = b P(a, b)
r r a
Let the point P(a, b) be reflected on X-axis. Then,
P(a, b) o P'(a, – b) X' T MX
-T
Then MOP' = 360° – T. O
sin (360° – T) = y-coordinate of P' = –b =– sinT
radius r
P'(a, -b)
cos (360° – T) = x-coordinate of P' = a = cosT
radius r
Y'
y-coordinate of P' –b
tan (360° – T) = x-coordinate of P' = a = – tanT
Similarly, we can show that -
sec (360° – T) = secT
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cot (360° – T) = – cotT
cosec (360° – T) = – cosecT
Hence, trigonometric ratios of (360° – T) and (– T) can be tabulated below:
sin(– T) = sin (360° –T) = – sinT
cos (–T) = cos (360° – T) = cosT
tan (–T) = tan(360° – T) = – tanT
cot (–T) = cot(360° – T) = – cotT
cosec (–T) = cosec(360° – T) = – coseT
sec (–T) = sec(360° – T) = secT
Note :
The trigonometric ratios of (360° – T) is same as that of trigonometric ratios of negative
angle (– T).
(i) Trigonometric Ratios of (360° + T) Y
In 360° + T, where T is an acute angle. 180O° T
The angle (360° + T) lies in the first quadrant. Y'
Hence the trigonometric ratios of (360° + T) and T are same. P(a, b)
X
Here, 360° + T = 1 complete turn + T X'
sinT = sin (360° + T) = b 180°
r
a
cosT = cos (360° + T) = r
tanT = tan (360° + T) = b
a
Alternative Methods of Some Trigonometric Ratios
Trigonometric Ratios of (180° – T)
Here, sin (180° – T) = sin {90° + (90° – T)} = cos (90° – T) = sinT
cos (180° – T) = cos {90° + (90° – T)} = – sin (90° – T) = – cosT
tan (180° – T) = sin(180° – T) = sinT = – tanT
cos(180° – T) – cosT
1 1
Also, cosec (180° – T) = sin(180° – T) = sinT = cosecT
Similarly,
cot (180° – T) = – cotT
sec(180° – T) = – secT
Trigonometric Ratios of (270° – T)
Here, sin (270° – T) = sin {90° + (180° – T)} = cos (180° – T) = – cosT
cos(270° – T) = cos{90° + (180° – T)} = – sin (180° – T) = – sinT
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tan(270° – T) = sin(270° – T) = – cosT = cotT
cos(270° – T) – sinT
Similarly,
cot(270° – T) = tanT
cosec (270° – T) = – secT
sec (270° – T) = – cosecT
Trigonometric Ratios of (270° + T)
Here, sin (270° + T) = sin {180° + (90° + T)} = – sin (90° + T) = – cosT
cos(270° + T) = cos {180° + (90 + T)} = – cos (90° + T) = – (– sinT) = sinT
tan(270° + T) = sin(270° + T) = –cosT = – cotT.
cos(270° + T) sinT
Similarly,
cosec (270° + T) = – secT
sec (270° + T) = cosecT
cot (270° + T) = – tanT
Tigonometric Ratios of (360° – T)
Here, sin (360° – T) = sin {180° + (180° – T)} = – sin (180° – T) = – sinT
cos(360° – T) = cos{180° + (180° – T)} = – cos(180° – T) = – (– cosT) = cosT
tan (360° – T) = sin(360° – T) = –sinT = – tanT
cos(360° – T) cosT
Similarly,
cosec (360° – T) = – cosecT
sec(360° – T) = secT
cot(360° – T) = – cotT
Signs of Trigonometric Ratios :
The signs of the trigonometric ratios of an angle depends on the quadrant in which the angle
lies.
Y
Second Quadrant First Quadrant
sine +
sine + cosine +
cosine – tan +
tan –
SA
(sine and cosec positive) (All ratios positive)
X' X
sine – O sine –
cosine +
cosine – C tan –
tan +
T
(tan and cot positive) (cos and sec positive)
Third Quadrant Fourth Quadrant
Y'
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The given table shows the signs of various trigonometric ratios in different quadrants:
Trigonometric Quadrants
ratios of angle T
First Second Third Fourth
sinT + –
cosT + +– +
tanT + –
cosecT + –– –
secT + +
cotT + –+ –
+–
––
–+
Trigonometric Ratios of Any Angle
If an angle can be expressed in the form of (n × 90° ± T), where n is an integer, even or odd.
The trigonometric ratios of the angle (n × 90° ± T), can be changed into the trigonometric
ratios of angle T and signs are taken according to quadrants in which they lie.
1. If 'n' is odd in the angle (n × 90° ± T), then the trigonometric ratios change from
sin to cos, cos to sin
tan to cot, cot to tan
cosec to sec, sec to cosec
Examples :
(i) sin (90° + T) = sin (1 × 90 + T) = cosT
(ii) cos(3 × 90° + T) = sinT
(iii) tan (270° + T) = tan (3 × 90° + T) = – cotT
2. If 'n' is even in the angle (n × 90° ± T), there is no change in the trigonometric ratios.
Examples :
(i) sin (180° – T) = sin(2 × 90 – T) = sinT
(ii) cos(180° – T) = sin(2 × 90° – T) = – cosT
(iii) tan(180° – T) = tan(2 × 90° – T) = – tanT
(iv) tan(180° + T) = tan(2 × 90° + T) = tanT
3. The signs of the trigonometric ratios of the angles (n × 90° ± T) is determined by
taking into considereation that in which quadrant that angle (n × 90° ± T) lies.
Examples :
(i) sin150° = sin(2 × 90° – 30°) = sin30° = 1 (It lies in the 2nd quadrant)
2
(ii) tan225° = tan (2 × 90° + 45°) = tan45° = 1 (It lies in the third quadrant)
(iii) cos315° = cos (4 × 90° – 45) = cos45° = 1 (It lies in the fourth quadrant)
2
1
(iv) cot420° = cot (4 × 90° + 60°) = cot60° = 3 (It lies in the first quadrant)
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(v) sin240° = sin (3 × 90° – 30°) = – cos30° = – 3 (It lies in the third quadrant)
2
3
or, sin240° = sin (2 × 90° + 60°) = – sin60° = – 2
Worked out Examples
Example 1. Find the values of the following :
Solution:
(a) sin120° (b) cos225° (c) tan300° (d) sin (–600°)
Example 2.
Solution: (a) Here, sin 120° (120° lies in the second quadrant)
= sin(90° + 30°) = cos30° = 3
2
(b) Here, cos225° (225° lies in the third quadrant)
= cos(180° + 45°) = – cos45° = – 1
2
(c) Here, tan300° (300° lies in the fourth quadrant)
= tan (3 × 90° + 30°) = – cot30° = – 3
(d) Here sin(– 600°) = – sin600°
= –sin (6 × 90° + 60°) = – (–sin60°) = 3
2
Prove that :
(a) sin75° = cos15°
(b) sin26° + tan70° = cos64° + cot20°
(c) tan81°. tan27°. tan63°. tan9° = 1
(a) Here,
LHS = sin75°
= sin (90° – 15°)
= cos15° = RHS proved.
(b) LHS = sin26° + tan70°
= sin (90° – 64°) + tan (90° – 20)
= cos64° + cot20° = RHS proved.
(c) LHS = tan81°. tan27°.tan63° + tan9°
= tan(90° – 9°). tan27°. tan(90° – 27°). tan9°
= cot9°.tan27°. cot27°.tan9°
= (cot9°. tan9°). (tan27°.cot27°)
= 1.1 = 1 = RHS proved.
Example 3. Prove that : Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
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(a) secT . cosec(90° – T) – tanT.cot (90° – T) = 1
(b) cos(90° + T). sec(–T).tan(180° – T) =– 1
sec(360° + T). sin(180° + T).cot(90° – T)
Solution: (a) LHS = secT. cosec(90° – T) – tanT.cot(90° – T)
Example 4.
Solution: = secT.secT – tanT.tanT = sec2T – tan2T
Example 5. = 1 = RHS proved.
Solutions:
(b) LHS = cos(90° + T). sec(–T).tan(180° – T) T)
sec(360° + T). sin(180 + T).cot(90° –
= (–sinT) secT(–tanT) = – 1 = RHS proved
secT.(–sinT).tanT
Find the numerical values of :
(a) sin70°.cos20° + sin20°.cos70°
(b) sin420°.cos390° + cos(–300°) sin(–330°)
(a) LHS = sin70°. cos20° + sin20°.cos70°
= sin(90° – 20°) cos20° + sin20°. cos(90° – 20°)
= cos20°.cos20° + sin20°.sin20°
= cos220° + sin220°
= sin220° + cos220°
= 1 ( sin2T + cos2T = 1)
= RHS proved.
(b) LHS = sin420°.cos390° + cos (–300°) sin (–330°)
= sin420°.cos390° + cos300° (–sin330°)
= sin(4×90°+60°).cos(4×90°+30°)–cos(4×90°–60°).sin(4×90°– 30°)
= sin60°.cos30° – cos60°(–sin30°)
= 3 . 3 + 1 . 1 = 3 + 1 = 3+1 = 4 =1
2 2 2 2 4 4 4 4
= RHS. proved.
Prove that :
(a) tan Sc . tan 3Sc . tan 5Sc . tan 7Sc = 1
8 8 8 8
Sc 3Sc 5Sc 7Sc
(b) cos 8 + cos 8 + cos 8 + cos 8 = 0
(a) LHS = tan Sc . tan 3Sc . tan (Sc – 3Sc ). tan (Sc – Sc )
8 8 8 8
Sc 3Sc 3Sc Sc
= tan 8 . tan 8 (–tan 8 ) . (– tan 8 )
= tan Sc . tan 3Sc (–tan 3Sc ). (–tan Sc )
8 8 8 8
Sc 3Sc
= tan2 8 . tan2 8
= tan2 Sc . tan2 ( Sc – Sc )
8 2 8
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= tan2 Sc . cot2 Sc
8 8
Sc Sc
= (tan 8 . cot 8 )2
=1 (tanT.cotT = 1)
= RHS proved.
Example 6. Solve for T, (0° ≤ T ≤ 90°) (b) tan2T = cot3T
Solution: (a) sin4T = cos2T
(a) Here, sin4T = cos2T
Example 7.
Solution: or, sin4T = sin(90° – 2T)
? 4T = 90° – 2T
Example 8. or, 6T = 90°
Solution: ? T = 15°
(b) Here, tan2T = cot3T
or, tan2T = tan(90° – 3T)
? 2T = 90° – 3T
or 5T = 90°
? T = 18°
Solve for x : cosec (90° + T) + x cos (–T). cot(90° +T) = sin(90° + T)
Here, cosec(90° + T) + x cos (–T).cot(90° + T) = sin(90° + T)
or, secT + xcosT. (– tanT) = cosT
or, –x cosT . sinT = cosT – secT
or, cosT
1
– x sinT = cosT – cosT
or, – x sinT = – 1 – cos2T
cosT
sin2T
or, x sinT = cosT
or, x= sin2T
cosT.sinT
or, x= sinT = tanT ? x = tanT
cosT
If A, B, and C are the angles of a triangle prove the following.
( )(b) tan C
(a) sin (A + B) = sinC A+B = cot 2
2
(a) Since A, B and C are angles of a triangle, we can write
A + B + C = Sc
or, A + B = Sc – C
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Taking sin on both sides,
sin (A + B) = sin (Sc – C)
or, sin (A + B) = sinC. Proved
(b) Here, A + B + C = Sc
or, A + B = Sc – C
or, A+B = Sc – C
2 2 2
Taking tan on both sides, we get,
tan ( A +B ) = tan ( Sc – C )
2 2 2
A +B C
? tan ( 2 ) = cot 2 Proved.
Exercise 11.1
Y
Very short Questions P(a, b)
1. (a) From the adjoining figure, MX
write the trigonometric ratios of X' r
sinT, cosT and tanT. O
(b) Write the trigonometric ratios of the
following in simplest form : Y'
(i) sin (90° – T) (ii) cos (90° + T) (iii) tan (180° – T) (iv) cot(90° – T)
(viii) tan (360° + T)
(v) tan(180° + T) (vi) sin(270° + T) (vii) cos(360° – T)
(iv) sec 150°
(c) Find the numerical values of the following : (viii) cos390°
(i) sin 135° (ii) tan225° (iii) cot150° (iv) sin 315°
(viii) cos570°
(v) cosec240° (vi) tan270° (vii) sin420°
(d) Find the numerical values of the following :
(i) sin (–420°) (ii) tan1020° (iii) cot(–960°)
(v) tan660° (vi) cosec (–570°) (vii) cos840°
Short Questions :
2. Prove the following:
(a) sin20°.cos70° + cos20°.sin70° = 1
(b) sin40°.cos50° + cos40°.sin50° = 1
(c) tan9°.tan27° = cot81°.cot63°
(d) tan16°.tan32° . tan74°.tan58° = 1
(e) cot9°.cot27°.cot45°. cot63° . cot 81° = 1
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3. Find the numerical values of the following :
(a) sin40° + cos40° + sin220° + cos140°
(b) sin2135° + cos2120° + tan2240°
(c) sin2180° + sin2150° +sin2135° + sin2120°
(d) sin2120° + cos2120° + sin2135° + tan2180°
(e) cos290° + cos2120° + cos2135° + cos2150° + cos2180°
(f) sin245° – 4sin260° + 2cos245° + cos2180°
4. Prove the following :
( ) ( ) ( ) ( )(a) sin 8 – sin 8 + cos 8 + cos 8 = 05Sc 3Sc
Sc 7Sc
Sc 3Sc 5Sc 7Sc
( ) ( ) ( ) ( )(b) cos 8 + cos 8 + cos 8 + cos 8 = 0
( ) ( ) ( ) ( )(c)
7Sc 5Sc 3Sc + sin2 Sc =
sin2 8 + sin2 8 + sin2 8 8
Sc
( ) ( ) ( ) ( )(d) cos2 8
3Sc 5Sc 7Sc =2
+ cos2 8 + cos2 8 + cos2 8
5. Find the value of T, (0° ≤ T ≤ 90°)
(a) sin2T = cosT (b) tan2T = cotT
(c) cos4T = sin2T (d) cos7T = sin3T
6. Show that:
(a) sin60°.cos150° + sin120°.cos330° = 0
(b) cos240°.cos210° + sin150°.cos300° = 3+1
4
3
(c) sin420° + sin30° + cos150° – sin240° + sin210° = 2
(d) sin30° – cos60° + tan45° + sin210° – cos240° + tan135° = 0
(e) sin150°.cos240° + cos(–330°). sin660° = –1
7. Prove the following :
(a) tan(90° – T). tan(270° + T) + cosecT.cosec(180° – T) = 1
(b) sin2(90° – T) + sin2T = 2
cos2T cos2(90°–
T
(c) sinT T)× cosT × tanT =1
cos(90° + sin(90°– T) cot(90°+ T)
(d) sin(180° – T) . cos(360° – T). cot (90° – T) = sinT
tan(180°– T) sin(– T) tan(90° + T)
(e) sin(180° – T) + tan(180° – T) + cos(180° – T) = – 1
sinT tanT cosT
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Long Questions
8. Prove the following geometrically:
(a) tan(180° – T) = – tanT (b) sin(180° + T) = – sinT
(c) tan(90° – T) = cotT (d) tan(– T) = – tanT
9. Prove the following geometrically:
(a) sin(270° – T) = – cosT (b) tan(270° + T) = – cot T
(c) cos(360° – T) = cosT
10. Solve the following equation for x:
(a) tan2225° – sin2120° = xsin45°.cos135° tan120°
(b) 3sin120° + xcos120°. tan135° = x cot330° tan150°
(c) x cot (270° + T).cot(90° + T) = tan(180° – T). tan(360° – T)cosec(90° – T).cosec(90° + T)
(d) tan(90° + T).cot(180° – T) + cosec(360°– T). cosecT = xcotT.tan(90° + T)
(e) sin2135° – xtan2150° = cos2150°
(f) sin120°.tan30° – xcos60° = tan45°.cos330°.sin120°
(g) x cotT.tan(90° + T) = tan(90° + T).cot(180° – T) + x sec(90° + T).cosecT
11. If A, B and C are the angles of a triangle prove the following:
( )(b) sin
(a) cos(A + B) = – cosC A+B = cos C
(c) tan (A + B) + tanC = 0 2 2
( )(d) tan
A+B . tan C =1
2 2
12. If A, B, C and D are the angles of a quadrilateral, then prove the following :
(a) sin (A + B) + sin (C + D) = 0 (b) cos(A+ B) = cos(B + D)
(c) tan(A +B) + tan(C + D) = 0 (d) cot(A + B) + cot(C + D) = 0
1. (b) (i) cosT (ii) – sinT (iii) – tanT (iv) tanT (v) tanT
(vii) cosT (viii) tanT
(vi) – cosT
(ii) 1
(c) (i) 1 (iii) – 3 (iv) – 2
2 3
(v) – 2 (vi) f (vii) 3 (viii) 3
3 2 2
(d) (i) – 3 (ii) – 3 (iii) – 1 (iv) – 1
2 3 2
(v) – 3 (vi) 2 (vii) – 1 (viii) – 3
2 2
15 3 3
3. (a) 0 (b) 4 (c) 2 (d) 2
(e) 5 (f) –3
2 2
5. (a) 30° (b) 30° (c) 15° (d) 9°
(c) sec2T (d) tan2T
10. (a) 1 (b) 3 3 (g) cot2T
23
3 1
(e) – 4 (f) – 2
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11.2 Trigonometric Ratios of compound Angles
Review
Group discuss the following questions:
(a) What are the possible compound angles of angles A and B ?
(b) Define a unit circle.
(c) How can you define trigonometric ratios in unit circle taking any point on the circle ?
Introduction
Let A and B be any two angles. Then the sum and the difference of these two angles are
written as A + B and A – B. These are called compound angles of A and B.
Definition : The sum or difference of two or more angles is called compound angle. Let A, B
and C be any three angles, then A + B, B + C, A – B, B – C, A + B + C, A – B – C, A – B + C
etc. are called compounded angles.
Trigonometric ratios in a unit circle
Let O be the origin and a unit circle is drawn taking O as Y
the centre. Let P(x, y) be any point on the circle and XOP = T.
Take a point S on X-axis such that OS = OP = 1 unit. Then P(x, y)
the coordinates of S is (1, 0). = (cos T, sinT)
r=1 y
TS
Now, sinT y-coordinte of P y y X' O xM X
? radius r 1
y = sinT
and cosT = x-coordinte of P = x = x
radius r 1
? x = cosT ? Y'
Hence P(x, y) = P(cosT, sinT)
(a) Trigonometric ratios of compound Angles (A + B) and (A – B)
Let us draw a unit circle taking origin O as the centre. Take a point S on X-axis such
that coordinates of S are (1, 0). Take P and Q two points on the circle such that
SOP = A, POQ = B. Let us take R on the Q(cos(A+B), sin(A+B)) Y B P(cosA, sinA)
circle below X-axis such that SOR = – B. A + X
Then, SOQ = A + B and X' BA S(1, 0)
O -B
POR = A + (– B) = A – B
From the above figure, the coordinates
of P, Q and R are writen as
P(cosA, sinA) R(cos(-B), sin(-B))
Q(cos (A + B)), (cos(A + B))
R(cos(–B), sin(–B)) = R(cosB, – sinB) Y'
Now, by using distance formula
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SQ2 = {cos(A + B) – 1}2 + {sin(A + B) – 0}2
= cos2(A + B) – 2cos(A + B) + 1 + sin2(A + B)
= sin2(A + B) + cos2(A + B) + 1 – 2cos(A + B)
= 1 + 1 – 2cos(A + B) = 2[1 – cos(A + B)]
Similarly,
PR2 = (cosB – cosA)2 + (–sinB – sinA)2
= cos2B – 2cosB.cosA + cos2A + sin2B + 2sinB.sinA + sin2A
= (sin2A + cos2B) + (sin2B + cos2B) – 2 cosA.cosB + 2sinA.SinB
= 1 + 1 – 2 (cosA.cosB – sinA.sinB)
= 2 – 2(cosA.cosB – sinA.sinB) = 2[1 – (cosA.cosB – sinA.sinB)]
Since, A + B = A + |–B|, we have,
SQ = PR (chords opposite to equal central angles)
or, SQ2 = PR2
or, 2[1 – cos(A + B)] = 2[1 – (cosA.cosB – sinA.sinB)]
or, 1 – cos(A + B) = 1 – (cosA.cosB – sinA.sinB)
? cos(A + B) = cosA.cosB – sinA.sinB ................. (1)
Then we get, cos(A + (–B)) = cosA.cos(–B) – sinA.sin(–B)
or, cos(A –B) = cosA.cosB + sinA.sinB
? cos(A – B) = cosA.cosB + sinA.sinB .................. (2)
Again, sin(A + B) = cos[90° – (A + B)]
= cos[(90° – A) – B]
= cos(90° – A). cosB + sin(90° – A). sinB
= sinA.cosB + cosA.sinB
? sin(A + B) = sinA.cosB + cosA.sinB ................. (3)
Let angle B be replaced by – B in (3),
Similarly,
sin(A + (–B)) = sinA.cos(–B) + cosA.sin(–B) = sinA.cosB – cosAsinB
? sin(A – B) = sinA.cosB – cosA.sinB ................. (4)
tan(A + B) = sin(A + B) = sinA.cosB + cosA.sinB
cos(A + B) cosA.cosB – sinA.sinB
(Dividing numerator and denominator by cosA.cosB)
= tanA + tanB
1 – tanA.tanB
? tan(A + B) = tanA + tanB ................ (5)
1 – tanA.tanB
Similarly, tan(A – B) = tanA – tanB .............. (6)
1+tanA.tanB
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cot(A + B) = cos(A + B) = cosA.cosB – sinA.sinB
sin(A + B) sinA.cosB + cosA.sinB
(Dividing numerator and denominator by sinA.sinB)
cot(A + B) = cotA.cot B – 1 ................... (7)
cotB + cotA
cotA.cotB + 1
Similarly, cot(A – B) = cotB – cotA
Trigonometric ratios of compound angles
1. sin(A + B) = sinA.cosB + cosA.sinB
2. cos(A + B) = cosA.cosB – sinA.sinB
3. tan(A + B) = tanA + tanB
1 – tanA.tanB
cotA.cotB – 1
4. cot(A + B) = cotB + cotA
5. sin(A – B) = sinA.cosB – cosA.sinB
6. cos(A – B) = cosA.cosB + sinA.sinB
7. tan(A + B) = tanA + tanB
1 – tanA.tanB
cotA.cotB +1
8. cot(A – B) = cotB – cotA
Some more results from trigonometric ratios of compound angles.
1. sin(A + B).sin(A – B) = cos2B – cos2A = sin2A – sin2B
Proof :
LHS = sin(A + B).sin(A – B)
= (sinA.cosB + cosA.sinB) (sinA.cosB – cosA.sinB)
= sin2A.cos2B – cos2A.sin2B
= (1 – cos2A).cos2B – cos2A(1 – cos2B)
= cos2B – cos2A.cos2B – cos2A + cos2A.cos2B
= cos2B – cos2A = middle term
Again, cos2B – cos2A = 1 – sin2B – 1 + sin2A = sin2A – sin2B
? sin(A + B).sin(A – B) = cos2B – cos2A = sin2A – sin2B = RHS. Proved.
2. cos(A + B). cos(A – B) = cos2A – sin2B = cos2B – sin2A. Proved.
Proof :
LHS = cos(A + B).cos(A – B)
= (cosA.cosB – sinA.sinB) (cosA.cosB + sinA.sinB)
= cos2A.cos2B – sin2A.sin2B
= cos2A(1 – sin2B) – (1 – cos2A)sin2B
= cos2A – cos2A.sin2B – sin2B + cos2A.sin2B
= cos2A – sin2B = middle side
Again, cos2B – cos2A = 1 – sin2B – 1 + sin2A = sin2A – sin2B
? sin(A + B).sin(A – B) = cos2A – sin2B = cos2B – sin2A = RHS
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3. tan (A + B).tan(A – B) = tan2A – tan2B
1 – tan2A.tan2B
Proof :
tanA + tanB
( ) ( )LHS = tan (A + B).tan(A – B) = 1 – tanA.tanB
tanA – tanB
1 + tanA.tanB
= tan2A – tan2B = RHS Proved.
1 – tan2A.tan2B
4. cot(A + B).cot(A – B) = cot2A . cot2B – 1
cot2B – cot2A
Proof :
LHS = cot(A + B).cot(A – B) = cotA.cotB – 1 . cotA.cotB+ 1
cotB + cotA cotB – cotA
cot2A.cot2B – 1
= cot2B – cot2A = RHS Proved.
5. sin(A + B + C) = sinA.cosB.cosC + cosA.sinB.cosC + cosA.cosB.sinC – sinA.sinB.sinC
Proof :
LHS = sin(A+B+C) = sin{(A + B) + C}
= sin(A +B).cosC + cos(A +B).sinC
= (sinA.cosB + cosA.sinB).cosC + (cosA.cosB – sinA.sinB) sinC
= sinA.cosB.cosC + cosA.sinB.cosC + cosA.cosB.sinC – sinA.sinB.sinC
= RHS Proved.
6. cos(A + B + C) = cosA.cosB.cosC –cosA.sinB.sinC – sinC.cosB.sinA – sinA.sinB.cosC
Proof :
LHS = cos(A+B+C)
= cos{(A+B)+C}
= cos(A+B).cosC – sin(A+B). sinC
= (cosA.cosB – sinA.sinB).cosC – (sinA.cosB + cosA.sinB).sinC
= cosA.cosB.cosC – sinA.sinB.cosC – sinA.cosB.sinC – cosA.sinB.sinC
= RHS. Proved.
7. tan(A+B+C) = tanA + tanB + tanC – tanA.tanB.tanC
1 – tanB.tanC – tanC.tanA – tanA.tanB
LHS = tan{(A + B) + C}
= tan(A + B) + tanC
1 – tan(A +B).tanC
( ( ) )=
tanA + tanB + tanC
1 – tanA.tanB
1– tanA + tanB . tanC
1 – tanA.tanB
= tanA + tanB + tanC – tanA.tanB.tanC
1 – tanA.tanB – tanA.tanC – tanB.tanC
= tanA + tanB + tanC – tanA.tanB.tanC = RHS. Proved.
1 – tanB.tanC – tanC.tanA – tanA.tanB
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Worked Out Examples
Example 1. Calculate the value of sin105° without using calculator or table.
Solution:
Example 2. Here, sin105° = sin(60° + 45°)
Solution:
= sin60° cos45° + cos60° sin45°
Example 3.
Solution: = 3 . 1 + 1 . 1
2 2 2 2
1
= 3 + 22 = 3+1
22 22
3 5
If sinA = 5 and cosB = 13 , find the value of (i) sin(A + B) (ii) cos(A – B)
Here, sinA = 3
5
cosA = 1 – sin2A , sinB = 1 – cos2B
( )= 1– 32 ( )= 1– 52
5 13
= 1 – 9 = 1 – 25
25 169
= 25 – 9 = 169 –25
25 = 169
= 4 144 = 12
5 169 13
(i) sin(A + B) = sinA.cosB + cosA . sinB
= 3 . 5 + 4 . 12
5 13 5 13
= 15 + 48 = 63
65 65 65
(ii) cos(A – B) = cosA.cosB + sinA.sinB
= 4 .153 + 3 . 12
5 5 13
20 36 56
= 65 + 65 = 65
Prove that tan75° – tan105° = 23
3 –1
Here
LHS = tan75° + tan105°
= tan(45°+30°) – tan(60° + 45)
= tan45° + tan30° – tan60° + tan45
1 – tan45° . tan30° 1 – tan60° . tan45°
1
1+ 3 – 3+1
1 – 3 .1
= 1
3
1–1.
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= 3+1 – 3+1
3 –1 1– 3
= 3+1 + 3–1
3 –1 3 –1
= 2 3 = RHS Proved.
3 –1
Example 4. If D + E = Sc, then prove that tanA + tanB + tanA.tanB = 1
Solution:
Here, D+E Sc
4
Example 5. Sc
Solution: ? tan(D + E) = tan 4
or, tanD + tanE = 1
1 – tanD.tanE
Example 6. or, tanD + tanE = 1 – tanD.tanE
Solution:
? tanD + tanE + tanD.tanE = 1 proved.
Example 7.
Solution: Prove that tan55° – tan35° = 2 tan20°
Here, 20° = 55° – 35°
? tan20° = tan(55° – 35°)
= tan55° – tan35°
1 + tan55°.tan35°
= tan55° – tan35°
1 + tan (90° – 35°). tan35°
= tan55° – tan35°
1 + cot35°.tan35°
= tan55° – tan35°
1+1
? tan20° = tan55° – tan35°
2
or, 2tan20° = tan55° – tan35°
? tan55° – tan35° = 2tan20° Proved.
Prove that cos(A + B) = cotA.cotB – 1
sinA.sinB
Here, cos(A + B) = cosA.cosB – sinA.sinB
sinA.sinB sinA.sinB
= cosA.cosB – sinA.sinB = cotA.cotB – 1
sinA.sinB sinA.sinB
= RHS Proved.
sin(A + 45°) = 1 (sinA + cosA)
2
Here,
LHS = sin(A + 45°)
= sinA.cos45 + cosA.sin45°
= s1i2nA(s1i2nA++cocsoAsA. 1)2
=
= RHS proved.
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Example 8. Prove that cos18° – sin18° = 2 sin27°
Solution: Here,
LHS = cos18° – sin18°
= cos(45 – 27°) – sin(45° – 27°)
= (cos45°.cos27° + sin45°.sin27°) – (sin45° . cos 27° – cos45° . sin27°)
= 1 cos27° + 1 sin27° – 1 cos27° + 1 sin27°
2 2 2 2
2
= 2 sin27°
= 2 sin27°
= RHS Proved.
Example 9. Prove that cos35° + sin35° = cot10°
Solution: cos35° – sin35°
Here,
LHS = cos35° + sin35°
cos35° – sin35°
= cos(45° – 10°) + sin(45° – 10°)
cos(45 – 10°) – sin(45° – 10°)
= cos45°.cos10° + sin45.sin10° + sin45°.cos10° – cos45°.sin10°
cos45°.cos10° + sin45.sin10° – sun45°.cos10° + cos45°.sin10°
1 cos10° + 1 sin10° + 1 cos10° – 1 sin10°
2 2 2 2
= 1
1 cos10° + 2 sin10° + 1 cos10° – 1 sin10°
2 2 2
2
2 cos10°
= 2 sin10° = cot10° = RHS Proved.
2
Example 10. If sinA = 1 and sinB = 1 , prove that A + B = Sc
Solution: 10 5 4
1 1
Here, sinA = 10 and sinB = 5
cosA = 1 – sin2A cosB = 1 – sin2B
= 1– 12 = 1 – 1
10 5
= 3 = 2
10 5
Now, cos(A + B) = cosA.cosB – sinA.sinB
= 3 . 2 – 1 . 1 = 6 – 1
10 5 10 5 50 50
6–1 1 5 1
= 50 = 50 = 52 = 2
? cos(A + B) = 1
or,
? cos (A + B) = 2 Sc
cos 4
A+B= Sc Proved.
4
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Example 11. If tanA = 1 and tanB = m, then show that A + B = Sc
Solution: m 2
1
Example 12. Here, tanA = m and tanB = m
Now, tan(A + B) = tanA + tanB
1 – tanA.tanB
1 +m 1 + m2
m m
= 1 = 1–1
1– m . m
1 + m2
m Sc
= 0 =∞ = tan 2
? tan(A + B) = tan Sc
2
Sc
? A+B = 2 proved.
If A + B = 45°, then prove that (1 + tanA) (1 + tanB) = 2
Solution: Here, A + B = 45°
? tan(A + B) = tan45°
or, tanA + tanB =1
1 – tanA.tanB
or, tanA + tanB = 1 – tanA.tanB
or, tanA + tanB + tanA.tanB = 1
Adding 1 on both sides,
tanA + tanA.tanB + tanB + 1 = 1 + 1
or, tanA(1 + tanB) + 1(1 + tanB) = 2
? (1 + tanB) (1 + tanA) = 2 Proved.
Example 13. Prove that tan20° + tan72° + tan88° = tan20°.tan72° + tan88°
Solution: Here, 92° = 20° + 72°
tan92° = tan(20° + 72°)
or, tan92° = tan20° + tan72°
1 – tan20°. tan72°
tan20° + tan72°
or, tan(180° – 88°) = 1 – tan20°. tan72°
or, –tan88° + tan20°.tan72°.tan88° = tan20° + tan72°
? tan20° + tan72° + tan88° = tan20°.tan72°.tan88° proved.
Example 14. If an angle T is divided into two parts D and E such that tanD : tanE = x : y
x–y
prove that sin(D – E) = x+y sinT
Solution: Since T is divided into two parts D and E,
We have, D + E = T
and tanD = x
tanE y
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By using componendo and dividendo If a = c
b d
tanD – tanE = x– y
tanD + tanE x+ y a – b c – d
sinD sinE a + b = c + d
cosE – cosE
x– y
or, sinD sinE = x+ y
cosD cosE
–
or, sinD.cosE – cosD.sinE = x–y
sinD.cosE + cosD.sinE x+y
or, sin(D – E) = x–y
sin(D + E) x+y
or, sin(D – E) = x–y
sinT x+y
x–y
? sin (D – E) = x+y sinT Proved.
Exercise 11.2
Very short Questions:
1. (a) Write the formula of the following:
(i) sin(A + B) (ii) sin(A – B) (iii) tan(A + B)
(iv) cot(A + B) (v) cos(A + B) (vi) cot(A –B)
(b) Find the value of the following using compound angle formula:
(i) cos15° (ii) sin15° (iii) sin75°
(iv) cos75° (v) tan75° (vi) tan105°
(vii) tan15° (viii) cos105° (ix) sin(–165°)
(x) tan(–195°)
Short Questions (xi) tan20° + tan25°
1 – tan20°. tan25°
2. Prove the following :
(a) sin75° + cos75° = 3 (b) sin15° + cos15° = 3
2 2
1
(c) tan15° + cot15° = 4 (d) sin105° + cos105° = 2
(e) sin75° – sin15° = 1
2
3. Prove the following :
(a) tan5° + tan40° + tan5°.tan40° = 1
(b) tan36° + tan9° + tan36°.tan9° = 1
(c) tan35° + tan10° = 1 – tan10°.tan35°
(d) tan17° + tan28° = 1 – tan17°.tan28°
(e) tan53° – tan8° = 1 + tan53°.tan8°
(f) tan25° + tan20° = 1 – tan25°.tan20°
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4. Prove the following :
(a) tan8T – tan5T – tan3T = tan8T.tan5T.tan3T.
(b) tan10° + tan70° + tan100° = tan10°.tan70° tan100°
(c) cot24° + cot23° + cot43° = cot24°.cot23°.cot43°
(d) tan80° + tan65° + tan35° = tan80° tan65° +tan35°
(e) tan15T – tan10T – tan5T = tan15T.tan10T.tan5T.
(f) tan(A – B) + tan(B – C) + tan(C – A) = tan(A – B). tan(B – C) tan(C – A).
5. (a) If sinD = 3 , and sinE = 12 , find the values of the following :
5 13
(i) sin(D – E) (ii) cos(D + E) (iii) tan(D + E)
(b) If cosT = 4 and cosI = 7 , find
5 52
(i) cos(T – I) (ii) sin(T – I) (iii) tan (T + I)
(c) If tanD = 1 and tanE = 1 , find the values of
4 5
(i) tan(D + E) (ii) cot(D + E) (iii) tan(D – E) (iv) cot(D – E)
6. Prove that :
(a) sin (T – 45°) = 1 (sinT – cosT)
2
(b) sin(45° + T) + cos(45° + T) = 2 cosT
(c) 2sin(T + 45°) . sin(T – 45°) = sin2T – cos2T
(d) sin2(T + 45°) + sin2(45° – T) = 1
(e) tan(45° – A) = cosA – sinA
cosA + sinA
(f) sin(x + 2)T. cos(x + 1) T – cos(x + 2)T.sin(x + 1)T = sinT
(g) sin5T. cos3T + cos5T.sin3T = sin9T.cosT – cos9T.sinT)
(h) 1 – 1 = cosec2A.
tanA tan2A
Long Questions
7. Prove that :
(a) cos9° + sin9° = tan54° (b) sin20° + cos20° = cot155°
cos9° – sin9° sin20° – cos20°
cos35° – sin35°
(c) cos35° + sin35° = tan10° (d) cos20° – cos70° = 2 sin25°
8. Prove that :
(a) 2tan50° = tan70° – tan20°
(b) 2tan20° = tan55° – tan35°
(c) 2tan70° = tan80° – tan10°
(d) 2tan10° = tan50° – tan40°
9. (a) If tanT = 5 , and tanI = 1 , then show that T + I = Sc
6 11 4
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(b) If cosT = 4 and cos I = 7 , prove that T + I = Sc
(c) If tanD = 5 52 4
m 1 that Sc
and tanE = 2m + 1 , then show D + E = 4 .
m+1
Sc
10. If D + E = 4 , then prove that :
(a) tanD + tanE + tanD.tanE = 1
(b) (1 + tanD) (1 + tanE) = 2
(c) cotE(cotD– ) – cotD = 1
(d) (cotD – 1) (cotE – 1) = 2
11. Prove that following
(a) sinA.sin(B – C) + sinB.sin(C – A) + sinC.sin(A – B) = 0
(b) sinA + sin(A + 120°) + sin(A – 120°) = 0
(c) cosA + cos(120° + A) + cos(120° – A) = 0
(d) (1 + tan21°) (1 + tan28°) (1 + tan24°). (1 + tan17°) = 4
12. If A + B + C = Sc and cosA = cosB.cosC, then prove that tanA = tanB + tanC
13. Prove that : sin(A – B) + sin(B – C) + sin(C – A) = 0
cosA.cosB cosB.cosC cosC.cosA
14. Prove the following :
(a) cos(A + B + C) = cosA.cosB.cosC (1 – tanA.tanB – tanB.tanC – tanC.tanA)
(b) tan(A + B + C) = tanA + tanB + tanC – tanA.tanB.tanC
1 – tanA.tanB – tanB.tanC – tanC.tanA
1 2
15. If tan(D + E) = 3 and tan (D – E) = 5 , then prove that
(a) tan2D = 11 (b) tan2E = –1
13 17
16. If an angle T is devided into two parts A and B such that tanA : tanB = x : y, show that
x–y
sin (A – B) = x+y sinT
17. If tanD = k tanE, then prove that sin(D – E) = k–1 sin(D + E)
k+1
18. If 2tanE + cotE = tanD, then prove that 2 tan (D – E) = cotE
19. If sinD + sinE = m and cosD + cosE = K then prove that cos (D – E) = 1 (m2 + n2 – 2).
2
20. If sin (T + I) = 2 sin(T – I), prove that tanT = 3 tanI.
1. (b) (i) 3+1 (ii) 3–1 (iii) 3+1 (iv) 3–1
22 22 22 22
1– 3
(v) (2 + 3) (vi) – (2 + 3) (vii) (2 – 3) (viii) 2 2
(ix) 1– 3 (x) 3 – 2 (xi) 1 5. (a) (i) – 33
2 2 65
16 63 31 17
(ii) – 65 (iii) – 16 (b) (i) 25 2 (ii) 25 2
(iii)1 (c) (i) 9 (ii) 19 (iii) 1 (iv) 21
19 9 21
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12Vector
12.0 Review
Group discuss the following questions :
(a) What is the height of your body?
(b) Why do we use bearing and scale drawing ?
(c) Define displacement and distance.
(d) Mr. Jhalak says to his friend, 'My house is 2000 metre away from our school. "Can his
friend locate the house of Jhalak ?
(e) Define physical quantities with examples.
12.1 Introduction
Which of the following are physical quantities ?
(i) Temperature (ii) Pressure (iii) Density (iv) Love
(v) Force (vi) Beauty (vii) Area (viii)Feelings
(ix) Displacement (x) Distance
Which of the above physical quantities need direction to specify them ?
The quantities which can be measured are called physical quantities Physical quantities are
divided into two categories.
(a) Scalar Quantities or scalars (b) Vector Quantities or vectors
(a) Scalar Quantities : The physical quantities which have only magnitude but no
direction are called scalar quantities or scalars. For examples; distance, mass, volume,
temperature, speed, density, area, time, height, etc. All these quantities can only be
determined by their magnitudes.
(b) Vector Quantities : The physical quantities which have both magnitude and direction are
called vector quantities or vectors. For examples; displacement, velocity, acceleration,
force, weight etc. These quantities involve both magnitude and direction.
Directed Line Segment
Let A and B be two points in the plane. The line segment B
joining these points A and B is denoted by AB. If point A is A
taken as initial point and B is as the terminating point, the
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line segment is called directed line segment and it is denoted by AoB (read as vector AB).
A directed line segment is a vector. It consists of three main parts, the initial point, the
terminal point (and ending point and the body (or length).
A vector is denoted by a single small letter or by a combination of two letters by arrow head
over it. For examples: AoB, PoQ, XoY, oa , ob , op , oq , etc.
Vector in Terms of Components
A vector can be expressed in terms of x-component and y-component.
For example : oa = 2 , ob = a etc.
4 b
(a) When the initial point of vector is at origin and the terminal point is at P(x, y)
Let O be the origin and P(x, y) be any point on the place. Join OP, PM is drawn
perpendicular on OX. Y
Then OM = x, PM = y P(x, y)
Then OoP can be written as a column vector. y
o x-component = x X' O x MX
OP = y-component y Y'
Q(x2, y2)
Example : If O(0, 0) and P(4, 5), R
Then, OoP = 4 NX
5
(b) When the initial point is at (x1, y1) and the terminating point is at (x2, y2)
Let P(x1, y2) and Q(x2, y2) be any two points on the plane.
PM and QN are drawn perpendicular to OX from P Y
and Q respectively. PR is drawn perpendicular P(x 1, y 1)
to QN.
Then, PR = MN = ON – OM
= x2 – x1 X' O M
QR = QN – RN = QN – PM Y'
= y2 – y1
Now,Vector PQ can be expressed as a column vector.
PoQ = x-component
y-component
= PR = x2 – x1
QR y2 – y1
Example : Let P(4, 5) and Q(7, 8) be two points.
Then PoQ = x2 – x1
y2 – y1
= 7–4 = 3
8–5 3
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Magnitude of a Vector
The magnitude of a vector is the distance between the initial and the terminal point of a
vector. The magnitude of a vector is also called absolute value or modulus of a vector. Its
value is always taken as positive. The magnitude of a vector AoB is denoted by |AoB|.
Let AoB = x , then its magnitude is given by
y
|AoB|
= (x-component)2+(y-component)2
= x2 + y2
Magnitudes of AoB is also denoted by AB.
o
Example : If P(4, 5) and Q(10 ,12), then find PQ and its magnitude.
Solution: Here, P(x1, y1) = P(4, 5) and Q(x2, y2) = Q(10, 12)
Then, PoQ = x2 – x1 = 10 – 4 = 6
y2 – y1 12 – 5 7
Now, magnitude of PQ is given by,
|PoQ| = (x-component)2+(y-component)2
= 62 + 72 = 36 + 49 = 85 units.
Direction of a Vector
The angle made by the vector with X-axis in the positive direction is called direction of a
vector. It is denoted by T.
In other words, the ratio of y-component to the x-component of a vector is called its direction.
Let OoP = x be a vector which makes an angle T with the positive X-axis in positive
y Y
direction. Then, draw PMAOX.
P(x, y)
PM y y-component
? tan T = OM = x = x-component
or, T = tan–1 y y
x
Hence, the direction of OoP is T = tan–1 y
x . X' T X
Ox
In the adjoining figure, M
the direction of MoN (T1) = 60° Y'
Y
the direction of NoM (T2) = 240° Q N
the direction of PoQ (T3) = 120° 120° 240° 60° X
the direction of QoP (T4) = 300°
X' O
300°
Example : Let OoP = 23 , its direction is given by P Y' M
2
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tan T = y-component = 2 = 1 = tan 30°
x-component 23 3
? T = 30°
Note :
If P(x1, y1) and Q(x2, y2) be two points of PoQ, its direction is given by
T = tan–1 y2 – y1 . To find the direction (T) of a vector in degree, the given table may be
x2 – x1
helpful. Use calculator to calculate angle (T).
SN x-component y-component Quadrant Value of T Example
1. + + 1st T = acute angle T = tan-1 1 = 26.57
2
2. – + 2nd T = 180° – acute T = tan-1 – 1 = 153.43
angle 2
3. – – 3rd T = 180° + acute T = tan-1 –1 = 206.57
angle –2
4. + – 4th T = 360° – acute T = tan-1 –1 = 333.43
angle 2
Note :
(i) If a vector is parallel to X-axis from left to right, then T = 0°,
e.g. T = tan–1 0 = 0°.
2
(ii) If a vector is parallel to X-axis from right to left, then T = 180°,
e.g. T = tan–1 0 = 180°.
–2
(iii) If a vector is at right angle to X-axis,
T = tan–1 1 = 90°
0
(iv) If a vectors parallel to Y-axis up to down, then T = 270°.
e.g. T = tan–1 –1 = 270°.
0
Example : Find the direction of vector OoP = (2, 0).
Solution:
Here OoP represents a vector parallel to X-axis.
So, tan T = 0
2
= 0 = tan 0°
? T = 0° is the required direction of OoP.
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Worked out Examples
Example 1. (a) Draw an arrow for each of vectors shown in the graph. Also write them
Solution: in column rector.
Example 2. (b) MN displices point M(3, 2) to N(8, 6). Find MoN and express it as a
Solution: column vector.
(a) From adjoint graph,
We observe that the initial point is at the origins.
For vector OP, we write an arrow from Y
O to P. 8
7
6 P(5, 5)
OoP = 5 5
i.e. 5 , as a column vector Q(-7, 4) 4
For vector OQ, we write an arrow 3
2
from O to Q, 1
OoQ = –7 X' -7 -6 -5 -4 -3 -2 -1-1O 1 2 3 4 5 X
4 -2
i.e. , as a column vector. -3
For vector OQ, we write an arrow from Y'
O to Q.
(b) In the adjoint figure is a vector. Y
Initial point (x1, y1) = M(3, 2)
Terminatiny point (x2, y2) = N(8, 6) 8
7
Now, for MoN, 6 N(8, 6)
x-component = x2 – x1= 8 – 3 = 5
y-component = y2 – y1 = 6 – 2 = 4 5
4
3
2
1 M(3, 2)
X' -3 -2 -1-O1 1 2 3 4 5 6 7 8 9 X
-2
-3
Expressing MoN as a column vector, we write Y'
MoN = 5 .
4
Find the magnitude and direction of OoP = 3 .
3
Here, OoP = 3
3
x-component = 3
y-component = 3
Magnitude of OoP is given by
|OoP| = (x-component)2+(y-component)2
= ( 3)2 + 32 = 3 + 9
= 12 = 2 3
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To find direction of OoP.
Let T be the angle made by OoP with positive direction of X-axis. Then,
tanT = y-component = 3 = 3 = tan60°
x-component 3
? T = 60°
(Since x-component and y-component both are positive, T lies in 1st quadrant)
Example 3. Find the direction of PoQ, where P(7, 8) and Q(6, 7).
Solution:
Here, (x1, y1) = P(7, 8)
(x2, y2) = Q(6, 7)
Now, x-component of PoQ = x2 – x1 = 6 – 7 = – 1
y-component of PoQ = 7 – 8 = – 1
Let T be the angle made by PoQ with positive direction of X-axis. Then,
tan T = y-component = –1 = 1 = tan(180° + 45°) = tan 225°
x-component –1
? T = 225°
(Since x-component and y-component both are negative, T lies on third quadrant.)
Exercise 12.1
Very Short Questions:
1. (a) Define vectors with examples.
(b) Define scalars with examples.
2. Classify the following physical quantities into vectors and scalars:
(a) Displacement (b) Distance (c) Speed
(d) Area (e) Force (f) Pressure
(g) Acceleration (h) Velocity (i) Weight
(j) Work (k) Temperature (l) Length
3. Draw arrow diagrams for the following vectors.
(a) OoP = 4 (b) OoA = –2 (c) oa = 7
8 8 –2
(d) oa = 2 (e) ob = 1 (f) op = 10
4 8 2
4. Find the magnitudes of the following vectors:
(a) op = 1 (b) oq = –2 (c) ot = 2 (d) oa = 2
3 4 5 4
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5. Find the direction of the following vectors:
(a) op = 1 (b) oq = 6 (c) oa = 3
1 23 3
(d) ob = 2 (e) ob = 4 (f) oc = 0
4 8 –3
Short Questions:
6. If AoB displaces A to B, find AoB and show it in a column vector.
(a) A(3, 4), B(10, 6) (b) A(–3, –4), B(2, 2)
(c) A(1, 2), B(4, 5) (d) A(4, 5), B(7, 8)
7. If MoN displaces M to N, find the magnitude and direction of MoN.
(a) M(4, 6), N(8, 10) (b) M(–2, –3), N(1, 0) (c) M(8, 4), N(5, 2)
8. If AoB displaces A to B and CoD displaces C to D, prove that |AoB| = |CoD|.
(a) A(–5, 4), B(1, –1) and C(4, 3), (10, 8) (b) A(5, 3), B(8, 1) and C(2, 0), D(–1, 2)
9. Let A(–2, 3), B(3, 5), C(x + 1, 4), and D(3, –1) be four points. If |AoB| = |CoD|, find the
value of x.
10. (a) A vector PoQ displaces a point P(7, 2) to Q(6, 3), find PoQ, |PoQ| and direction of PoQ.
(b) A vector MoN displaces M(6, 1) to N(8, 5). Find MoN, |MoN| and direction of MoN.
Project Work
11. List the physical quantities which are used in our society. Classify them into scalars
and vectors with reasons.
2. (a) Vector (b) Scalar (c) Scalar (d) Scalar
(e) Vector (f) Scalar (g) Vector (h) Vector
(i) Vector (j) Scalar (k) Scalar (l) Scalar
4. (a) 10 units (b) 20 units (c) 29 units (d) 20 units
5. (a) 45° (b) 30° (c) 45° (d) 63.43° (e) 64.43° (f) 270°
6. (a) AoB = 7 (b) AoB = 5 (c) AoB = 3 (d) AoB = 3
2 6 3 3
7. (a) 4 2, 45° (b) 3 2, 45° (c) 13, 213.69° 9. x = 4, 0
10. (a) PoQ = –1 , 2, 135° (b) MoN = 2 ,2 5, 63.43°
1 4
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12.2 Types of Vectors
(a) Row Vector Y Q(9, 14)
In the adjoining figure MoN displaces M(2, 1) to 14
13
12
N(11, 10). Then, the horizontal displacement of 11 5 N(11, 10)
MN is 9 units and its vertical displacement is 9 10
units. 9 8
8
P(1, 9)
Hence, we write 7
MoN = (9, 9) 6
5 9
4
3
2 M(2, 1)
Again, PoQ displaces P(1, 9) to Q(9, 14). 1 9
X' O 1 2 3 4 5 6 7 8 9 10 11 X
Y'
Then, the horizontal displacement of PoQ is 8 units
and the vertical displacement of PoQ is 5 units.
Here, PoQ = (8, 5) is called row vector.
? PoQ = (8, 5)
Definition: If x-component and y-component of a vector are written in a horizontal order
enclosing in pair of brackets ( ), then the vector is called a row vector.
For example: oa = (4, 5), PoQ = (8, 4).
(b) Column Vector
From above graph, the vectors MoN and PoQ can be expressed in the form of MoN = 9
and PoQ = 8 9
5 .
Definition: If x-component and y-component of a vector are written in a vertical order
enclosing in a pair of brackets ( ), then the vector is called a column vector.
Example : oa = 4 , ob = –4 , etc.
5 5
(c) Position Vector Y
P(3, 3)
Let O be the origin and P(3, 3) be a point on the plane.
Then OP is joined. The vector OoP is called position
vector.
? OoP = 3 X' X
3
Definition: A vector whose initial point is at the origin and O
Y'
the terminal point is other than the origin is known as a
position vector.
(d) Zero/Null Vector
Vectors oPP, oPP, O = (0, 0), etc. are called zero vector or null vector.
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Definition: A vector whose magnitude is zero is called zero vector. oa = 0
0
x and y components both are zero in null vector.
(e) Unit Vector
Let oa = 1 , ob = 1 , 1 be two vectors. Let us find their magnitudes.
0 2 2
|oa | = 12 + 02 = 1 + 0 = 1 unit
|ob | = 1 2 1 2 = 1 + 1 = 1 units
2 2 2 2
+
Each of above vector has magnitude one. Such vectors are called unit vectors.
Definition: A vector whose magnitude is unity or 1 is called a unit vector. A vector oa is
called unit vector of |oa | = 1.
Example : Let = 3 , 4
5 5
then its magnitude is |oa | = 9 16 9 + 16 25
3 2 4 2 = 25 + 25 = 25 = 25 = 1 unit.
5 5
+
Unit Vectors oi and oj
The unit vector along positive X-axis is denoted by oi and its Y
x-component is unity and y-component is 0.
So, oi = 1 = (1, 0) X' B(0, 1) X
0
In the figure, OoA = oi = oi + 0 O A(1, 0)
Y'
The unit vector along position Y-axis is denoted by oj
oj = (0, 1) = 0 + oj
In the figure,
OoA = (1, 0) = oi
OoB = (0, 1) = oj
To find unit vector along a given vector:
A vector AoB divided by its magnitude |AoB| is known as a unit vector along the direction of
AoB.
It is denoted by A^B. AoB
|AoB|
Hence a unit vector along AoB =
i.e. A^B = AoB
|AoB|
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For example: Let AoB = (4, 3)
|AoB| = 42 + 32 = 16 + 9 = 25 = 5
Now, unit vector along AoB = AoB
|AoB|
i.e. A^B = (4, 3) = 4 , 3
5 5 5
(f) Negative Vector
Let us observe the two vectors AoB and PoQ as shown in the graph. From adjoint graph,
we have Y
AoB = 7 12 B(10, 11)
6 11
10
|AoB| = 72 + 62 9
8 P(14, 8)
= 49 + 36 = 85 units 7
6
5
PoQ = –7 4 A(3, 5)
|PoQ| –6
3
2
= ( –7)2 + (–6)2 1 Q(7, 2)
X' O 1 2 3 4 5 6 7 8 9 101112131415 X
= 49 + 36 = 85 units Y'
Here two vectors are parallel to each other and they have same magnitude but they
have opposite signs.
Hence, AoB and PoQ are negative vector of each other. It means AoB is negative vector of
PoQ and PoQ is negative vector of AoB.
We write, AoB = – PoQ
or, PoQ = – AoB
Definition: Two vectors oa and ob are said to be negative of each other if they have the same
magnitudes but opposite in direction. Then we write oa = – ob .
Negative vector of AoB is written as BoA.
If AoB = 4 , then BoA = – 4 = –4
5 5 –5
(g) Like Vectors B
D
In the adjoining figure AoB and CoD are two parallel vectors
with same direction. They are called like parallel vectors.
Definition: Two vectors are said to be like parallel vectors if they A
have the same direction regardless of their magnitudes. Like
vectors are also like parallel vectors. C
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