Trigonometry

Page 184 7A + 5A 7A - 5A
2 .a . Prove that 2sin 2 cos 2 - sin6A
sin2A + sin5A - sinA = 7A + 5A 7A - 5A
cos2A +cos5A + cosA = tan2A 2 cos 2 .cos 2 - cos6A

sin2A + sin5A - sinA 2sin6A. cosA - sin6A
LHS, cos2A +cos5A + cosA = 2cos6A.cosA -cos6A

5A + A 5A - A sin6A( 2cosA - 1)
sin2A + 2cos 2 sin 2 = cos6A(2cosA -1)
= 5A + A 5A - A
cos2A + 2 cos 2 .cos 2 = tan6A
sin2A + 2 cos3A.sin2A
= cos2A + 2 cos3A.cos2A RHS Proved
sin2A( 1 + 2cos3A)
= cos2A( 1 + 2cos3A) sinA.sin2A + sin3A.sin6A
= tan2A sinA.cos2A + sin3A cos6A = tan5A
RHS Proved.
2. d. sinA.sin2A + sin3A.sin6A
sin7A- sin6A +sin5A LHS sinA.cos2A + sin3A cos6A
cos7A -cos6A + cos5A = tan6A
2 sinA.sin2A + 2sin3A.sin6A
sin7A- sin6A +sin5A = 2 sinA.cos2A +2 sin3A cos6A
LHS, cos7A -cos6A + cos5A
cos( A-2A) - cos(A + 2A)+ cos(3A - 6A) - cos(3A + 6A)
sin7A + sin5A - sin6A = sin(A+2A)+sin(A -2A) + sin(3A+ 6A) +sin(3A -6A)
= cos7A + cos5A - cos6A
cosA-cos3A + cos3A - cos9A
= sin3A -sinA + sin9A - sin3A

A +9A 9A -A
cosA -cos9A 2 sin 2 sin 2
= sin9A -sinA =
9A +A 9A -A
2 cos 2 sin 2

sin5A Conditional Trigonometric Identities
= cos5A = tan5A RHS Proved. If A + B + C = 180 or c
A + B + C = 180
sin2A +sin4A +sin6A +sin8A or, A + B = 180 - C
cos2A +cos4A + cos6A +cos8A = tan5A Taking sin on both sides
Sin(A+B) = sin( 180- C)
sin2A +sin4A +sin6A +sin8A or, Sin(A+B) = SinC
LHS, cos2A +cos4A + cos6A +cos8A Taking cos on both sides
A + B = 180 - C
sin2A +sin8A +sin4A +sin6A or, cos(A + B) = cos (180-C)
= cos2A +cos8A + cos4A +cos6A or,cos( A +B) = -cosC
A + B + C = 180
2A +8A 2A -8A 4A +6A 4A -6A A + B = 180 - C
2sin 2 cos 2 +2sin 2 cos 2 divide by 2 on both sides
= 2A +8A 2A -8A 4A +6A 4A -6A A + B 180- C
2cos 2 cos 2 +2cos 2 cos 2
2sin5A.cos3A + 2 sin5A.cosA 2= 2
= 2cos5A.cos3A + 2 cos5A.cosA A B 180 C
2sin5A(cos3A + cosA) 2 +2 = 2 -2
= 2cos5A(cos3A +cosA) Taking sin on both sides
= tan5A. Proved

AB 180 C
Sin(2 + 2 ) = sin( 2 - 2 )

AB C
or, Sin( 2 + 2 ) = sin ( 90 - 2 )

AB C
or, Sin( 2 + 2 ) = cos 2

Taking cos on both sides

AB 180 C
cos( 2 + 2 ) = cos( 2 - 2 )

AB C
or, cos( 2 + 2 ) = cos ( 90 - 2 )

AB C
or, cos( 2 + 2 ) = sin 2

190 Page

1.a

or, tan2A + tan2B =-tan2C + tan2A.tan2B.tan2C
or, tan2A + tan2B + tan2C = tan2A.tan2B.tan2C proved.

If A + B +C = 180 then d. Prove that Tan( A+B-C) + tan( B + C -A) + tan(C +A - B)
tan2A + tan2B + tan2C = tan2A tan2B tan2C = tan2A.tan2B.tan2C
we have A + B + C = 180
or, A + B = 180 - C LHS , Tan( A+B-C) + tan( B + C -A) + tan(C +A - B)
multipling by 2 on both sides
or, 2(A + B) = 2( 180 - C) tan( 180 - C - C) + tan( 180- A -A) + tan( 180 - B -B)
or, 2A + 2B = 360 - 2C = tan2A.tan2B.tan2C
taking tan on both sides
tan( 2A + 2B) = tan( 360- 2C) or, tan( 180-2C) + tan( 180 - 2A) + tan( 180 -2B) =
tan2A.tan2B.tan2C
tan2A + tan2B
or, 1 - tan2A.tan2B = -tan2C or, - tan2C - tan2A - tan2B = tan2A.tan2B.tan2C
or, tan2A + tan2B = -tan2C( 1 - tan2A.tan2B)
or, -( tan2C + tan2A + tan2B ) = tan2A.tan2B.tan2C

or. tan2A +tan2B + tan2C = - tan2A.tan2B .tan2C

If A + B +C = 180 then or, cos(A + B) = cos (180-C)
sin2A - Sin2B + sin 2C = 4 cosA.sinB.cosC or,cos( A +B) = -cosC
we have sin2A - Sin2B + sin 2C
A + B + C = 180
or, A + B = 180 - C 2A + 2 B 2A - 2B
Taking sin on both sides = 2cos 2 sin 2 + sin2C
Sin(A+B) = sin( 180- C) = 2 cos(A+B) .sin(A-B) + sin2C
or, Sin(A+B) = SinC = 2(-cosC) .sin( A-B) + 2sinC.cosC
Taking cos on both sides = 2 cosC{ -sin(A-B) + sinC}
A + B = 180 - C = 2cosC{sinC - sin(A-B) }
= 2cosC { sin(A +B) - sin(A-B)}
= 2cosC.2cosA.sinB
= 4 cosAsinBcosC.
2. a . If A + B + C = 180 prove that

ABC
sinA + sinB + sinC = 4 cos 2 cos 2 cos 2
Solution: we have
A + B + C = 180
or, A + B = 180 - C
dividing by 2 on both sides

A + B 180- C C A-B C
2= 2 = 2 cos 2 ( cos 2 + sin2 )

A B 180 C C A-B AB
2 +2 = 2 -2 = 2 cos 2 { cos( 2 )+ cos( 2 + 2 )}

Taking sin on both sides C AB AB
= 2 cos2 { cos( 2 + 2 ) + cos( 2 -2 )}
AB 180 C
Sin(2 + 2 ) = sin( 2 - 2 )
C AB ABC
= 2cos2 2 cos 2 cos2 = 4 cos 2 cos 2 cos 2 Proved
AB C
or, Sin( 2 + 2 ) = sin ( 90 - 2 )
4. If A + B + C = 180 prove that

AB C Sin2A + sin2B + sin2C = 2 + 2cosAcosB .cosC
or, Sin( 2 + 2 ) = cos 2
we have A + B + C = 180

Taking cos on both sides or, A + B = 180 - C

AB 180 C
cos( 2 + 2 ) = cos( 2 - 2 ) Taking sin on both sides

AB C Sin(A+B) = sin( 180- C)

or, cos( 2 + 2 ) = cos ( 90 - 2 ) or, Sin(A+B) = SinC

AB C Taking cos on both sides
or, cos( 2 + 2 ) = sin 2

LHS sinA + sinB + sinC A + B = 180 - C

A+B A-B or, cos(A + B) = cos (180-C)
= 2 sin 2 .cos 2 + sinC
or,cos( A +B) = -cosC

C A-B CC Solution, Sin2A + sin2B + sin2C
= 2 cos 2 . cos 2 + 2sin2 cos2

1- cos2A 1- cos2B + sin2C or, cosA.cosB + sinA.sinB = -cosC
=2 +2
or, cosA.cosB + cosC = -sinA.sinB

1 cos2A 1 cos2B + sin2C squaring on both sides
=2- 2 +2 -2

1 ( cosA.cosB +cosC)2 = sin2A.sin2B
= 1 - 2 ( cos2A + cos2B)
+ sin2C or, (cosA.cosB)2 + 2cosA.CosB.cosC + cos2C = sin2A.sin2B

= 1 - 1 ( 2 cos 2A + 2B .cos 2A - 2B )+ sin2C or, cos2A.cos2B + 2cosA.CosB.cosC + cos2C = sin2A.sin2B
2 2 2

= 1 - cos(A +B) .cos(A-B) + sin2C or, ( 1 - sin2A) ( 1- sin2B) + 2cosA.CosB.cosC + ( 1- sin2C)
= sin2A. sin2B

= 1 - ( -cosC) . cos(A-B) + 1 - cos2C or, 1 - sin2B - sin2A + sin2A.sin2B + 2cosA.CosB.cosC
+ 1 - sin2C = sin2A.sin2B
= 2 +cosC.cos(A-B) - cos2C

= 2 + cosC{ cos(A-B) -cosC} or, 2 + 2cosA.CosB.cosC = sin2A.sin2B + sin2A + sin2B +
sin2C - sin2A.sin2B
= 2 + cosC{cos(A-B) + cos(A + B) }

= 2 + cosC . 2cosA.cosB or, 2 + 2cosA.CosB.cosC = sin2A + sin2B + sin2C
Proved.

= 2 + 2cosA.cosB.cosC

Alternative method

sin2A + sin2B + sin2C = 2 + 2cosAcosB .cosC If A + B + C = 180 prove that
A + B + C = 180

A + B = 180 - C sin2A2 + sin2B2 + sin2C2 ABC
= 1 - 2 sin 2 sin 2 sin 2

taking both cos on both sides A + B + C = 180

cos( A +B) = cos( 180 -C) or, A + B = 180 - C

dividing by 2 on both sides or, ( 1 - sin2A2 ) ( 1 - sin2B2 ) = sin2A2 sin2B2 +2 A B sin
sin 2 sin 2
A + B 180- C
2= 2 C + sin2 C
2 2

A B 180 C or, 1 - sin2B2 - sin2A2 + sin2A2 sin2B2 = sin2A2 sin2B2 +2 A
2 +2 = 2 -2 sin 2

Taking cos on both sides B C + sin2 C
sin 2 sin 2 2

AB 180 C
cos( 2 + 2 ) = cos( 2 - 2 )
or, AB C = sin2A2 + sin2B2 + sin2 C + sin2A2
1 - 2 sin 2 sin2 sin 2 2

AB C sin2B2 - sin2A2 sin2B2
or, cos( 2 + 2 ) = cos ( 90 - 2 )

AB C AB C = sin2A2 + sin2B2 C
or, cos( 2 + 2 ) = sin 2 or, 1 - 2 sin 2 sin2 sin 2 2
+ sin2 Proved .

A B AB C AB
or, cos 2 cos2 - sin 2 sin2 = sin 2 6. a . If A + B + C = 180 then proved that cos 2 +cos2 + cos
C B + C C+ A A+B
A B AB C 2 = 4 cos 4 cos 4 cos 4 =
or, cos 2 cos 2 = sin 2 sin2 + sin 2

Squaring on both sides. -A -B -C
4 cos 4 cos 4 cos 4
or,( cos A cos B )2 = A B + sin C )2 Solution
2 2 (sin 2 sin2 2

or, cos2A2 cos2B2 = sin2A2 sin2B2 +2 A B C + sin2 C A + B + C = 180
sin 2 sin 2 sin 2 2
or, A + B = 180 - C

dividing by 4 on both sides

A + B 180- C A+B A-B C+ A+B
4= 4 = 2cos 4 .cos 4 + 2cos 4 cos 4

A B 180 C A+B A-B C+
4 +4 = 4 -4 = 2cos 4 ( cos 4 + cos 4 )

Taking cos on both sides A-B C+ A-B C+
4 +4 4 -4
AB 180 C A+B cos 2
= 2cos 4 . 2 cos
cos( 4 + 4 ) = cos( 4 - 4 ) 2

Taking sin on both sides A+B A -B +C+ A -B -(C+)
= 4cos 4 cos 4x2 cos 4x2

AB 180 C
sin( 4 + 4 ) = sin ( 4 - 4 ) A+B A -B +C+A+B+C A -B -(C+A+B+C)
= 4cos 4 cos cos
8 8

A+B 2A +2C A -B -C-A-B-C
=4cos 4 cos 8 .cos
AB C 8
cos 2 +cos2 + cos 2
A+B A +C -2B-2C
AB C = 4cos 4 cos 4 cos 8
= cos 2 + cos2 + cos 2 + cos 2
A+B A +C B +C
A B AB C  C = 4 cos 4 cos 4 cos 4
2 +2 2 -2 2 +2 2 -2
= 2cos 2 .cos 2 + 2cos 2 cos 2 A+B B +C A +C
= 4 cos 4 cos 4 cos 4

A+B A-B C+ C- -C -A -B
=2cos 4 .cos 4 + 2cos 4 cos 4 = 4 cos 4 cos 4 .cos 4

A+B A-B C+ -C
= 2cos 4 .cos 4 + 2cos 4 cos 4

sin(A+B+C) = sin180

or,sin{ (A+B)+C} = sin180

or, sin(A+B).cosC + cos(A+B).sinC = 0

or,(sinA.cosB + cosA.sinB) cosC + ( cosA.cosB -sinA.sinB)

sinC = 0

or, sinA.cosB.cosC + cosA.sinB.cosC + cosA.cosB.sinC -

sinA.sinB.sinC = 0

or, sinA.cosB.cosC + cosA.sinB.cosC + cosA.cosB.sinC =

1. If A + B + C = 180 prove that sinA.sinB.sinC
Sin A .cosB . cosC + cosA .sinB.cosC + cosA.cosB .sinC =
sinA.sinB.sinC If A + B + C = 180 prove that cosAsinBsinC +
Here A + B + C = 180
A + B = 180 - C cosBsinCsinA + cosCsinAsinB = 1 + cosAcosBcosC
taking cos and sin on both sides
cos(A + B ) = cos( 180 -C ) = - cosC here, A + B + C = 180
sin(A + B ) = sin ( 180 - C ) = sinC
From LHS taking cos on both sides.
Sin A .cosB . cosC + cosA .sinB.cosC + cosA.cosB.sinC
= cosC( SinA.cosB + cosA.sinB) + cosA.cosB.sinC cos (A+B+C) = cos180
= cosC sin(A + B) + cosA.cosB .sinC
= cosC sinC + cosA.cosB .sinC or, cos{(A+B) + C} = -1
= sinC ( cosC + cosA.cosB )
= sinC { -cos( A +B) + cosAcosB} or, cos(A + B) .cosC - sin(A+B) .sinC = -1
= sinC { - (cosA.cosB -sinA.sinB ) + cosA.cosB}
= sinC{- cosA.cosB + sinA.sinB + cosA.cosB} or, (cosAcosB - sinA.sinB) cosC - (sinA.cosB +cosA.sinB )
= sinC.sinA.sinB
= sinA.sinB.sinC Proved sinC = -1

Alternative method or, cosAcosBcosC -sinAsinBcosC - sinAcosBsinC -
A + B + C = 180
taking sin on both sides. cosAsinBsinC = -1

or, 1 + cosAcosBcosC = sinAsinBcosC + sinAcosBsinC +

cosAsinBsinC

 cosAsinBsinC + cosBsinCsinA + cosCsinAsinB = 1 +

cosAcosBcosC RHS proved.

If A + B + C = 180 prove that sin(B + C - A ) +sin (C +A -

B) + sin(A + B - C) = 4 sinA.sinB.sinC

here,

A + B + C = 180

A + B = 180 - C

B + C = 180 - A

C + A = 180 - B

Taking sin and cos on A + B = 180 - C

Sin(A + B ) = sin( 180 - C ) = sinC If A + B + C = 180 prove that cos( B + C -A ) + Cos (C + A -
cos( A + B ) = sin ( 180 - C ) = -cosC B ) + cos (A + B - C )
From question
sin(B + C - A ) +sin (C +A - B) + sin(A + B - C) = 1 + 4cosAcosBcosC
= sin( 180 - A - A ) + sin ( 180 - B -B) + sin( 180 - C - C) If A + B + C = 1800 Prove that
= sin ( 180 - 2A) + sin(180 - 2B) + sin( 180 - 2C)
= sin2A + sin2B + sin2C B-C
sin( B + 2C) + sin (C + 2A) + sin( A + 2B) = 4 sin 2 sin
2A+2B 2A-2B
= 2sin 2 cos 2 + sin2C C - A A -B
2 sin 2
= 2sin(A+ B) .cos(A- B ) + sin2C
= 2 sinC.cos(A-B) + 2sinC.cosC
= 2sinC { cos(A- B) + cosC}
= 2 sinC { cos(A- B ) - cos(A+B) }
= 2sinC .2sinA.sinB
= 4sinA.sinB.sinC RHS proved

A + B + C = 180

A + B = 180 - C

B + C = 180 - A

C + A = 180 - B

LHS ,sin( B + 2C) + sin (C + 2A) + sin( A + 2B)

= sin( B + C + C) + sin (C + A + A) + sin( A + B +B)

= sin( 180 - A + C) + sin (180 -B + A) + sin( 180 - C +B)

= sin{ 180 + ( C - A)} + sin { 180 + ( A -B)} + sin{ 180 + (B

- C)}

= - sin ( C - A) - sin( A -B) - sin(B - C)

= sin(A - C) + sin( B - A) - sin(B - C)

(A -C) + (B-A) (A -C) - (B-A)
= 2 sin cos 2 - sin(B - C)
2

A - C +B -A A - C -B +A
= 2 sin 2 cos 2 - sin(B - C)

B -C 2A -B-C B -C B -C
= 2sin 2 cos 2 - 2 sin 2 cos 2

B -C 2A -B-C B -C 2cosA.sinA + 2cosB.sinB + 2cosC.sinC
= 2sin 2 ( cos 2 - cos 2 ) = 2sinA.sinB.sinC

2A -B-C B -C sin2A +sin2B + sin2C
B -C 2 + 2 = 2sinA.sinB.sinC
= 2 sin 2 × - 2 sin
2 taking
sin2A + sin2B + sin2C
2A -B-C B -C
2 -2 2A +2B 2A -2B
= 2 sin 2 cos 2 +sin2C
sin 2 = 2 sin(A + B) .cos(A - B) + sin2C
= 2 sinC.cos(A-B) + 2sinC.cosC
B -C 2A -B -C + B -C 2A -B -C -B +C = 2sinC{ cos(A - B) + cosC}
= -4 sin 2 sin 4 sin 4 = 2 sinC{ cos(A-B) - cos(A +B)}
= 2sinC .2sinA.sinB
B -C 2A - 2C 2A - 2B = 4sinA.sinB.sinC
= - 4 sin 2 sin 4 sin 4
now from above
B -C A-C A-B sin2A +sin2B + sin2C
= - 4 sin 2 sin 2 sin 2 = 2sinA.sinB.sinC
4sinA.sinB .sinC
B -C C- A A-B = 2sinA.sinB.sinC
= 4 sin 2 sin 2 sin 2 Proved . = 2 Proved

If A + B + C = 1800 Prove that

cosA cosB cosC
sinB.sinC + sinC.sinA + sinA.sinB = 2

Here A + B + C = 180

A + B = 180 - C

taking cos and sin on both sides

cos(A + B ) = cos( 180 -C ) = - cosC

sin(A + B ) = sin ( 180 - C ) = sinC

LHS

cosA cosB cosC
sinB.sinC + sinC.sinA + sinA.sinB

cosA.sinA + cosB.sinB + cosC.sinC
= sinA.sinB.sinC

Squences and series. 2. d = b-a
n-1

1, 2, 3,4,5,6,.......sequence 3. t1 = a

1 + 2 + 3 + 4 .................. series 4. t2 = a + 1d

2 , 4 , 6 , 8 ................ 5. t3 = a + 2d

first term = a = t1 = 2

second term = t2 = 4 1 , 2 , 3 are in AP
here 2 is called mean between 1 and 3 .
third term = t3 = 6 a,b,c are in AP
here b is called mean between a and c .
common difference = d = t2-t1 = t3 - t2 1,2,3,4,5,6,7,8,9,10 are in AP .
here 2,3,4,5,6,7,8 and 9 are called means between 1 and 10.
or, d = 4 - 2 = 6 - 4 If a,b,c are in AP then prove that

or, d = 2 = 2 c+a
b = mean = 2
A P = Arithmetic progession.

t1 = a = a +0.d

t2 = a + d = a +1d

t3 = a + d + d = a + 2d

t4 = a + d + d +d = a + 3d

t5 = a + d+d+d+d = a +4d t1 = a, t2 = b and t3 = c
d = t2 - t1 = t3- t2
.................................................... or, b - a = c - b
or, b - a = c - b
...................................................... or, b + b = c + a
or, 2b = c + a
t9 = a +8d
c+a
t10 = a +9d or, b = 2

t100 = a +99d = a + ( 100-1)d c+a
b = mean = 2
tn = a +(n-1)d
a+b
1.tn = a+ (n-1)d If a,m and b are in AP the proved that m = 2 .
we have a,m,b are in AP
when tn = last term = b let a = t1

tn = a+ (n-1)d m = t2
or, b = a + (n-1)d b = t3
or, b -a = (n-1)d

b-a
or, n-1 = d = common difference for terms.

t2 - t1 = t3 - t2 b-a
or, m -a = b -m therefore, comman difference for mean ,d = nm + 1
or, m + m = a + b
or, 2m = a + b b-a
therefore, comman difference for term = d = nt -1
a+b
therefore m = 2 b -a
d = n-1 here, n = no of term
Relation between no of terms and no of means.
no of terms = no of means + 2 b -a
d = n+1 here, n = no of mean
or, nt = nm + 2
common difference for term Sn = sum of nth term of AP
tn = a + ( n-1)d n
let tn = b = last term
or ,b = a + ( n-1) d Sn = 2 { 2a +(n-1)d} when n,a & d are given.
or , b -a = ( n-1) d n

b -a Sn = 2 { a + a + (n-1)d}
or, n-1 = d = common difference for term. here n = no of n
term = nt
no of term = no of mean + 2 Sn = 2 ( a + tn)
nt = nm + 2 n
we know that
Sn = 2 ( a + b) when a,n & b( last term)
b -a 9
d = n-1
b-a 5a. If first term (a) = 18 and common diffrence ( d) = - 2
or, d = nt -1 then find t9.
Given ,
b-a a= 18
or , d = nm + 2 -1 here, nt = nm + 2
9
b-a d=- 2
therefore, d = nm + 1
9
t9 = a + 8d= 18 + 8(-2 ) = 18 -36=-18

6. c . If the 17th term of AP is 6 and its common difference d = t2 - t1 = t3 - t2
5 d = 20 - 17 = 23 -20 = 3
4 , find its first term. tn = a + (n - 1 )d
or, 44 = 17 + ( n-1) 3
5 or, 44 -17 = 3(n-1)
Given d = 4 or, 27 = 3 ( n-1)
t17 = a + 16d = 6
27
5 or, 3 = n - 1
or, a + 16 x 4 = 6 or, 9 = n -1
or, a + 20 = 6 or, n = 10, therefore 44 is the t10 of given AS.
or, a = 6-20 = -14
7. c. If 2k + 8 ,6k and 8k -1 are in AP, find the value of k 10. Find the 15th term of the sequence
and the three term. 3,6,9,12,........................
Given,
2k + 8 ,6k and 8k -1 are in AP 49-5n
i.e, 12. b. tn = 2 is an general term of AP.
t2 - t1 = t3 - t2 when n = 1
or, 6k - ( 2k +8) = 8k -1 - 6k
or, 6k - 2k -8 = 2k -1 49-5x 1
or, 4k -2k = -1+8 then, t1 = 2 = 22

7 t2 =
or, 2k = 7therefore , k = 2 t3 =
d = t2 - t1 = t3 - t2
7 therefor given tn is the general term.
t1=2k + 8 = 2x 2 + 8 = 7 + 8 = 15
3.a. In the arithmetic sequence ,5t5 = 9t9,then prove that t14
7 = 0.
t2 = 6k = 6 x 2 = 3 x 7 = 21 given
5t5 = 9t9
7 or, 5 ( a + 4d) = 9 ( a + 8d)
t3 = 8k -1= 8x2 - 1 = 28 - 1 = 27 or, 5a + 20d = 9a + 72d
8.a. Is 44 a term of the AS 17, 20, 23.......? or, 0 = 9a -5a + 72d - 20d
Given or, 0 = 4a + 52d
t1= 17
t2= 20
t3 = 23

or,0 = 4 (a + 13d) 5. a. Find the equation of the line perpendicular bisector
0 of the line segment joining the points (2, - 3) and (- 6, 1).

or,4 = a +13d D
or, 0 = t14 proved .
A ( 2 , -3) B ( -6 , 1)
10. a .
a =3 C ( m, n)
d=6-3=9-6=3
t15 = ? Let C( m,n) be the midpoint of AB,
n= 15
t15 = a + 14d = 3 + 14 x 3 = 3 + 42 = 45 here, A( 2,-3) = A ( x1,y1)
14 .B
page 28 B( -6,1) = B ( x2,y2)
a=7
by using midpoint formula

x = x1 + x2 = 2 - 6 -4 = -2
2 2 =2

y = y1 + y2 = -3 +1 -2 = -1
2 2 =2

midpointC( m,n) = C( -2,-1)

slope of AB(m1) = y2 - y1 = 1+3 4 -1
x2 - x1 -6-2 = -8 =2

let m2 be slope of CD, by using perpendicular ,

m1.m2= -1

-1
or, 2 .m2 = -1

or, m2 = 2

now equation of line CD which is passing through the Group D ( 1 x 5 = 5)
pointC( -2,-1) with slope (m2) = 2 is 5. Transformation
y -y1 = m2( x - x1)
or, y - ( -1) = 2 { x - ( -2 ) } 1.Find the limit of 1.1,1.01,1.001,1.0001........ = 1
or, y + 1 = 2 ( x + 2 ) 2.Find the limit of 2.9,2.99,2.999,2.9999,.......
or, y + 1 = 2x + 4 =3
or, 0 = 2x - y +3 is the required equation of CD. 3.Find the limit of 6.5,6.05,6.005,6.0005,.........
=6
Subjective question 4.Write the formula of cot3A.
Opt. Math Class - 10 5.Write the formula of tan3A.
Group A ( 1 x 3 = 3 ) 6.Write the formula of sin3A.
1. a. Limit 7.Write the formula of cos3A
8. Find the image of P(x,y) under the reflection on x =0.
b. trigonometry 9. Find the image of P(x,y) under the reflection on y =0.
c. Transformation 10. Find the image of P(x,y) under the reflection on
Group B ( 2x 2 = 4)
2. a. function y= k.
b. Sequence and series 11. Find the image of P(x,y) under the reflection on
Group C( 2x 4 = 8)
3. MD from mean and median x =h.
4. SD

12. Find the image of P(x,y) under the reflection on 21. Find the image of the p(x,y) under the enlargement
y = x. with scale k and cetre ( 0,0).

13. Find the image of P(x,y) under the reflection on 22. Find the image of the p(x,y) under the enlargement
y = -x. with scale k and cetre ( a,b).

14. Find the image of P(x,y) under the rotation through
+900 about origin.

15.Find the image of P(x,y) under the rotation through
-900 about origin.

16. Find the image of P(x,y) under the rotation through
-900 about any point( a,b).

17. Find the image of P(x,y) under the rotation through
+900 about any point( m,n).

18. Find the image of P(x,y) under the rotation through
1800 about any point( 0,0).

19. Find the image of P(x,y) under the rotation through
1800 about any point( p,q).

20. Find the image of P(x,y) under the translation vector
a

T=( b )