Mathematics Tricks (2nd Edition) C-O

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T20+1= 23C20 . a23-20b20
T21= 23!/20!. 3! a3b20
T21= 23.22.21/6 a3b20
T21= 23.11.7 a3b20
T21= 1771 a3b20 (Answer)
Q#4: The total number of terms in the expansion of (1-x+x4)4 is:

A) 5 B) 15 (Correct) C)10 D) 8

Solution: As we know that: )
Total number of terms in (x1+x2+…..+xr)n is n+r-1Cr-1

whereas r= no. of terms
And also for trinomial expansion total terms is ( )(

In this case:
M#1: (1-x+x4)4
As r=3 so Total number of terms 4+3-1C3-1= 6C2=15 (Answer)
M#2: Number of terms in trinomial expansion= ( )( )= ( )=15 (Ans)

Q#5: The 6th term from the end in the expansion − is:

A) 16x B) 77x C) D) (Correct)

Solution: The pth term from the end in the expansion is (n-p+1)th term

from beginning.
so (11-5+1)th 7thso r=6

Now. Using the general term formula:

T6+1 =11C6 −

T6+1 =11C6 ()

T6+1 =11!/5!.6! −

T6+1 =11.10.9.8.7.6!/5.4.3.2.1.6! −

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T6+1 =11.3.2.7 −

T6+1 =11.7 −

T6+1 =(− )(Answer)
Q#6: The coefficient of the term involving x5 in the expansion of

− ? A) 15309 B) -15309 C) -15309/8(Correct) D) 15309/8

Explanation:
Trick:
The coefficient of the term involving xm in the expansion of
(xp+(a/xq))n is Tr+1=nCr an-r.br whereas r=

Apply the trick: r= ( ) = =5
So T5+1= 10C5 (x2)10-5.(-3/x2)5= -15309/8 (Answer)

Q#7: The term free from Z in the expansion of + is:

A) 5/4 B) ¼ C) 9/4 D) 4/5

Explanation:
Trick:
The coefficient of the term involving xm in the expansion of (xp+(a/xq))n
is Tr+1=nCr an-r.br whereas r=

Here in this case m=0 as independent term/term free from
variable/constant term contain x0 always so that:

() =2

r= / =

T2+1= 10C2 (z/3)4.(3/2z2)2 =5/4 (Answer)

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Q#8: The term free from t in (4t+5/t)8 is?

Explanation:
Using trick: r= np/p+q r= 1(8)/1+1 4
so Tr+1= T5  5th term (Answer)
Q#9: Total number of terms in the expansion of (x+z)50+(x-z)50 are?

Explanation:
Trick: Total terms of (x+y)n+(x-y)n are:

1) If n=odd then total terms are (n+1)/2

2) If n=even then total terms are (n/2)+1

In this case: (50/2)+1= 26 (Answer)
Q#10:The coefficient of x5 in the expansion of (1+x2) (1+x)4 is?

Explanation:
Using binomial theorem on (1+x)4
=(1+x2)(1+4x+6x2+4x3+x4)
=1+4x+7x2+8x3+4x5+x6
So coefficient of x5 is 4 (Answer)

Q#11:
The coefficient of terms x100 in the expansion of (1-x)-3 is?

Explanation:

Using the formulas:
(r+1)th term of (1-x)-3 is ( )( ) ( )( ) = 5151

Q#12:
If the coefficient of (2r+4)th and (r-2)th term in the expansion of (1+x)18

are equal?

Explanation:

Using the property”
nCx=nCy  n=x+y
so: 2r-3+r-3=183r=18r=6 (Answer)

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Q#13:The sum of coefficient of last 3 last term in the expansion of (8-
3x)1/2?

Explanation: The last term of the binomial series does not exist so can’t

be determined.
Q#14: If the coefficient of x2 and x3 in the expansion of (3+ax)9 are same

then a=?

Explanation:

Using the formula:
Tr+1= nCr. an-rbr
Tr+1= 9Cr (3)9-r(ax)r
For x2 T3= 78732(ax)2
For x3 T4=61236(ax)3

As the coefficient are equal so that:
78732(ax)2=61236(ax)3
 a= 9/7 (Answer)
Q#15:The greatest positive integer divides 11n-10n-1 is?

Explanation:

Put n=1 it gives the notation n=0
Put n=2 it gives the 112-10(2)-1= 121-20-1=121-21=100

which means greatest integer is 100.. (Answer)

Q#16:
If S(n)= n2-n+41 is a proposition then which of the following is not prime

A) S(1) B) S(1) and S(10) C) S(10) D) S(41)

Explanation:

On putting the value of n from the given options and satisfies the

condition.

so option D is satisfying.
S(41)= (41)2-(41)(41)

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S(41)=412=1681

As we know that prime number is one which has at most 2 divisor i.e

itself and 1.

In this case there are three divisor of 1681 i.e 1681 ,41 and 1 (Answer)

Q#17:
If a3+b3+c3+….+(2n-1)3=n2(2n2-1) then value of (a,b,c)=?

Explanation:
Put n=1 in both (2n-1)3 and n2(2n2-1) and get the answer.
As in this case (2(1)-1)3=1 so n2(2n2-1) is also 1

Put n=2 in both notation so:
(2(2)-1)3=27 so n2(2n2-1)=1+27=28

Put n=3 in both notation so:
(2(3)-1)3=125 so n2(2n2-1)=1+27+125=153
so (1,27,125)= (a3,b3,c3)

taking cube root on both sides so:

(1,3,5)=(a,b,c) (Answer)

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CHAPTER 9-14 TRIGONOMETRY

PUT AND CHECK TRICK:-

Suppose you have to solve an identity or any trigonometric equation but you
don’t remember the formula then there is a very easy method to guess the
correct answer.
Just put any value from the domain of a trigonometric function in the question,
you will get some value (Note that). Now put the same value in all the given
options and check for which option your answer matches. It will be the answer.

Example:-

2sinθ =?

a. cos 3θ b. sin 3θ c. tan 3θ d. cot 3θ

Now put θ=30° in 2sinθ =2sin30°=2(1/2)= 1 Note it…

Now put θ=30° in all options

Cos3θ =cos3(30°)=cos90°= 0
sin3θ =sin3(30°)=sin90°= 1
tan3θ =sin3θ /cos3θ =1/0=Undefined
cot3θ =cos3θ /sin3θ =0/1=0

So, right option is b. 2sinθ =sin3θ
Note:
This trick is very helpful. Especially in 9.3, 9.4 and 10.3, 10.4 also in chapter 13, 14.
Where a long equation come to solve practice it. But remember the domain and
try it that a function should not become undefined as tan90 and cot0 etc.

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Otherwise if in equation there come only sine and cos functions, solutions with
90,0 is more easy. Practice it maximum.

SOME USEFUL RESULTS TO REMEMBER:-

1. tan x + tan(180-x) = 0
2. cot x + cot(180-x) = 0
3. sin x – sin(180-x) = 0
4. cos x – sin(90-x) = 0
5. sin x – cos (90-x) = 0
6. cos x + cos(180-x) = 0
7. sin²α+cos²β=1 if α+β= 90°
8. sin nπ = 0
9. cos nπ= (-1)n n Є Z
10. sin(2n+1) = (-1)n

11. cos(2n+1) = 0

12. tan nπ= 0
13. cot nπ= ∞

14. tan(2n+1) = ∞

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15. cot(2n+1) = 0

SHORTCUT TO FIND DOMAIN, RANGE,
PERIOD & Frequency

Let,
Y = A (Trigonometric Function) Bx+C -------------------------------(1)

Then,
 Domain =
 Range = A (Range of Function)
 Period =
 Frequency =
 Phase shift=

e.g.,
y = 3 sin 4x+9

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Comparing it with eq. 1,

A=3, B=4, C=9

We know the Domain of sinx is R(All real numbers), Range is [-1,1] and
Period is 2π. So, for y=3sin4x

 Domain = = R

 Range = 3 [-1,1] = [-3,3]
 Period = =

 Frequency = =

 Phase Shift= -9/4

Quick Guessing Points for the Period of any function:

 Constant Function is a function with no fundamental period.
e.g; 1 has no period.

 If f(x) is periodic with period T, then 1/f(x) and ( ) are also
periodic with the same period.
e.g; √2 2 + 3 has π as period.

 sinx, cosx, secx, and cosecx are periodic functions with period 2π.
 tanx and cotx are periodic functions with period π.
 |sinx|,|cosx|,|secx|,|cscx|,|cotx| and |tanx| are periodic

functions with period π.
 Sinnx, cosnx, secnx and csecnx are periodic function with period 2π

when n is odd and π when n is even.
 tannx and cotnx are periodic functions with period π.

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Range of Function of type af(x)+b if f(x) is a trigonometric
Function.

 asinx+b=[b-a,b+a]
e.g range of 2sinx+5 is [5-2,5+2]=> [3,7]

 acosx+b= [b-a,b+a]
 atanx+b= (-∞,∞)
 acotx+b= (-∞,∞)
 asecx+b= (-∞,b-a]U[a+b,+∞)

e.g 2secx+9= (-∞,7]U[11,+∞)
 acsecx+b=(-∞,b-a]U[a+b,+∞)

Range of sin2x, cos2x, tan2x, sec2x, csec2x and cot2x can be got
by eliminating negative portion of range of function without
square.
e.g sin2x has range [-1,1] now eliminate the negative part of
range of trigonometric function and get the real range of
sin2x=[1].

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--Tricks on Clock Related Problem:

1) An accurate clock shows 8 o'clock in the morning. Through how many degrees

will the hour hand rotate when the clock shows 2 o'clock in the afternoon?

(a) 144 (b)150
(c) 168 (d)180
Sol - 8 o'click in morning - 2 o'clock = 6 hours.
using trick,
In 1 minute, an hour hand covers 1/2 degree and 1 hour = 60 minute. So, an hour hand
will cover (1/2) x 60 = 30 degrees in 1 hour. In total, we have to find the degrees movement of an

Hour clock in 6 hour so, it should be :
30 x 6= 180

(2) How many times are the hands of a clock at right angle in a day?
(a)22 (b) 24
(c)44 (d) 48
Sol - In 12 hours, they are at right angles 22 times so in 24 hours, they will at right angle at 44
times.

(3) How many times do the hands of a clock coincide in a day?

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(a) 22 (b)24
(c) 44 (d) 48
Sol- Have a look on above show data, it is clearly written than hands of a clock overlap each
other 22 times in a day so the answer will be 22.

(4) A clock is started at noon. By 10 minutes past 5, the hour hand has turned
through:

(a) 145 degrees (b) 150 degrees
(c)155 degrees (d)160 degrees
Sol- 10 minutes past 5 means :: 5 : 10. It can also be said that it is 5 hours and 10 minutes.
For Minute Hand
(i) 1 min = 1/2 degree so 10 min = 5 degrees
(ii) 1 hour = 60 minute = 30 degree
(iii) 5 hour = 5 x 30 degree = 150 degree
But we have to find degree subtended by 5 Hours and 10 Minutes = 150 + 5 = 155 degrees

(5) The angle between the minute hand and the other hour hand of a clock when the time is 8:30
is
(a) 80 degrees (b) 75 degrees
(c) 60 degrees (d) 105 degrees

Sol - In this kind of question, use the formula which has been written above.

Degree required = (11/2) *30 - 30*8
= 15*11 - 240
=165 -240
= 75 degree

Note - Here please note that, you can get answer in negative form but never consider negative
answer because it applies mod value

Maximum and Minimum Value of special case:
acosx+bsinx+c

Maximum value: c+ √ +
Minimum value: c - √ +

 Maximum value of acosx+bsinx-c =√ + −

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-Maximum and Minimum values of trigonometric expressions.

Type -1

Example - (1) Find the maximum and Minimum Value of 3 sin 2x + 4 cos 2x
Sol- Here the 4> 3 so

 Maximum Value = 4
 Minimum Value = 3

Example - (2) Find the maximum and Minimum Value of 5 sin 2x + 3 cos 2x
Sol - Here 5>3

 Maximum Value = 5
 Minimum Value = 3

Type -2

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Example - (1) Find the Maximum and Minimum Value of 3 sin x + 4 cos x

Sol- If you find the question of this kind, apply the above formulae.

 Maximum Value = √ 9 + 16 = √ 25 = 5
 Minimum Value = - √ 9 + 16 = - √ 25 = - 5

(2) Find the Maximum and Minimum Value of 3 sin x + 2 cos x

Sol- If you find the question of this kind, apply the above formulae.

 Maximum Value = √ 9 + 4 = √ 13
 Minimum Value = - √ 9 + 4 = - √ 13

Type-3

In case of sec2x, cosec2x, cot2x and tan2x, we cannot find the maximum value because they
can have infinity as their maximum value. So in question containing these trigonometric
identities, you will be asked to find the minimum values only. The typical question forms are
listed below:

Example- (1) Find the Minimum value of 3 sin 2x + 4 cosec 2x
sol - this equation is a typical example of our type-3 so apply the formula 2√ab so,
 Minimum Value = 2√ 3 x 4= 2√ 12
Example (2) Find the Minimum value of 9 cos 2x + 2 sec 2x
sol - this equation is a typical example of our type-3 so apply the formula 2√ab so,

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 Minimum Value = 2√ 9 x 2= 2√ 18
Example (3) -Find the Minimum value of 8 tan 2x + 7 cot 2x
sol - this equation is a typical example of our type-3 so apply the formula 2√ab so,

 Minimum Value = 2√ 8 x 7= 2√ 56

Type-4

In this type, we will give you the explanation of question which are different from
type-1, type-2 and type-3. If you find this kind of questions, you will have to convert

these questions into type-1,2 or 3 by using trigonometric formulas
Example - Find the Minimum Value of Sec 2x + cosec 2x

Sol - 1 + tan 2x + cosec 2x -------------------------------------------- (Sec 2x = 1 + tan 2x)

= 1+ tan 2x + 1 + cot 2x ------------------------------------------------ (cosec 2x = 1 + cot 2x )

=2 + tan 2x + cot 2x---------------------------------------------------apply type-3 formula

=2 + 2 √ 1 x 1

=2+2
=4 (Answer)

Important Notes & Short Tricks on Height & Distance

Angle of Elevation: Let AB be a tower/pillar/shell/minar/pole etc.) standing at
any point C on the level ground is viewing at A.

The angle , which the line AC makes with the horizontal line BC is called angle
of elevation .so angle ACB is angle of elevation.

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Angle of Depression: If observer is at Q and is viewing an object R on the
ground , then angle between PQ and QR is the angle of depression .so angle

PQR is angle of depression.

Numerically angle of elevation is equal to the angle of depression.
Both the angles are measured with the horizontal.

1. The thread of a kite is 120 m long and it is making 30° angular
elevation with the ground .What is the height of the kite?

Solution:

Sin 30° = h/120
1/2 = h/120
h = 60m

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2. A tree bent by the wind .The top of the tree meets the ground at an
angle of 60°.If the distance between the top of the foot be 8 m then
what was the height of the tree?

Solution:

tan 60° = x/8
√3 = x/8
x = 8 √3
y cos 60° = 8/y
1/2 = 8/y
y = 16
therefore height of the tree = x+y
= 8√3+16
= 8(√3+2)

3. The angle of elevation of the top of a tower from a point on the
ground is 30° . On walking 100m towards the tower the angle of
elevation changes to 60° . Find the height of the tower.

Solution:

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In right triangle ABD,
tan 60° = h/x
√3 x = h
x = h/√3
Again , in right triangle ABC ,
tan 30 = h/x+100
1/√3 = h/x+100
√3 h = x+100
√3 h = h/√3 + 100
√3 h – h/√3 =100
3 h - h/√3 =100
2 h = 100√3
h = 50√3
By short trick:
d = h (cot Ɵ1 - cot Ɵ2)
h = 100/(√3-1/√3) = 100*√3/2 = 50√3
Ɵ1 = small angle
Ɵ2 = large angle
d = distance between two places
h = height

4. From the top of a temple near a river the angles of depression of
both the banks of river are 45° & 30°. If the height of the temple is
100 m then find out the width of the river.

Solution:

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tan 45° = AB/BD
1 = 100/BD
BD = 100
tan 30 ° = AB/BC
1/√3 = 100/BC
BC = 100 √3
Width of the river , CD = BC - BD = 100 (√3-1)
When height of tower is 1 m then width of river is √3-1
Since height of tower is 100 m
Therefore ,
Width of river is 100(√3-1)m
By short trick:
Same formula can be used in this question too i.e.
d= h (cot Ɵ1 - cot Ɵ2)

5. The angle of elevation of the top of a tower from a point is 30 °. On
walking 40 m towards the tower the angle changes to 45°.Find the
height of the tower?

Solution:

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tan 45° = AB/BD
1 = AB/1
Therefore AB = 1
tan 30° = AB/BC =>1/√3 = 1/BC
therefore BC= √3
Now CD =√3-1 m and height of tower is 1 m
1 m = 1/√3-1
Therefore 40 m = 1/√3-1.40 = 40/√3-1
= 20 (√3+1)m
By trick:
40 = h(√3-1)

H = 40/(√3-1) = 20 (√3+1)m
Here are some ratio figure which you have to remember

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Important short trick are :

Note: only when the sum of angle i.e

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Some Important question are as follows:

Example 1:The angle of elevation of the top of a tower at a distance of 500 m
from its foot is 30°. The height of tower is :

(a)

(b)
(c)500

(d)
Ans. (d)

Short trick:
Solve it with ratio , as the angle of elevation is 30° then ratio between P: B: H

is 1:√3:2 so √3= 500 then 1= 500/√3 and height is equal to
Example 2: The banks of a river are parallel. A swimmer starts from a point on
one of the banks and swims in a straight line inclined to the bank at 450 and
reaches the opposite bank at a point 20 m from the point opposite to the
starting point. The breadth of the river is :
(a) 20 m
(b) 28.28 m
(c) 14.14 m
(d) 40 m
Ans. (c) 14.14 m
Solution:
Let A be the starting point and B, the end point of the swimmer. Then AB = 20m
&

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Short Method;
AS the angle of elevation is 45° then the ratio of P: B : H i.e. 1:1:√2
here √2 =20 then 1 =20/√2
Question 3: A man from the top a 50m high tower, sees a car moving towards
the tower at an angle of depression of 300. After some time, the angle of
depression becomes 600. The distance (in m) travelled by the car during this
time is –
(a)

(b)

(c)
(d)
Ans. (c)
Solution:

AB = AC – BC

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Example 4:A person standing on the bank of a river observes that the angle of
elevation of the top of a tree on the opposite side of the bank is 600. When he
moves 50m away from the bank, the angle of elevation becomes 300. The
height of the tree and width of river respectively are :

(a)

(b)

(c)
(d) None of these
Answer: c)
Solution:

Ratio value original value

height of the tree= h (ratio value = )=
and width of the river = x (ratio value = 1) = 25 m

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Example 5: From the top of a pillar of height 80 m the angle of elevation and
depression of the top and bottom of another pillar are 300 and 450respectively.
The height of second pillar (in metre) is:
(a) m

(b)

(c)

(d)
Answer: (c)
Solution:
Let AB and CD are pillars.
Let DE = h

In

Required height

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Some Important Results: =1
If α+β=90*
1) sinα=cos β
2) cosα=sinβ
3) tanα=cotβ
4) cotα=tanβ
5) secα=cosecβ
6) cosecα=secβ
Important Points:

If α+β=90*

1) sinα/cos β= cosα/sinβ=1
2) tanα/cotβ=cotα/tanβ= 1
3) secα/cosecβ= cosecα/secβ=1
-Results:
If A+B+C=π ,then:
1) tanA+tanB+tanC=tanA.tanB.tanC
2) cotA.cotB+cotB.cotC+cotC.cotA=1
3) . + . + .

4) + + = . .

-Domains of trigonometric Function:

Function Domain

Y=sinx R

Y=cosx R

Y=tanx R-{(2n+1)π/2 , nԑZ}

Y=cotx R-{nπ; nԑZ}

Y=secx R-{(2n+1)π/2 , nԑZ}

Y=cosecx R-{nπ ; nԑZ}

-Domain of Identities:

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Identity Domain
Sin2x+cos2x=1 R
1+tan2x=sec2x R-{(2n+1)π/2 , nԑZ}
1+cot2x=cosec2x R-{nπ; nԑZ}
-Domain of Reciprocal:
Function Domain
R-{(2n+1)π/2 , nԑZ}
Y=

Y= R-{nπ ; nԑZ}

Y= . x≠ , nԑZ

-Application of Trigonometry:
Types of triangles:

Name of triangle Sides Angles

Scalene 3 sides are 3 angles are
different different

Isosceles 2 sides are same 2 angles are same

Equilateral 3 sides are same 3 angles are same

Obtuse May be scalene or 1 angle obtuse
isosceles but not
equilateral

Right May be scalene or 1 angle is 90*
isosceles but not
equilateral

Oblique May be scalene or No angle of 90*
isosceles but not
equilateral

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Acute May be scalene, 3 angles are
isosceles or acute(less than
equilateral 90*)

-Types of Angles:

1) Acute angle (Less than 90*)
2) Obtuse angle (greater than 90*)

3) Straight angle (180*)
4) Reflex angle (360*)
5) Right angle (90*)

6) Complementary angle (Sum of two angles is 90*)
7) Supplementary angle (sum of two angles is 180*)
Trigonometric Triplets Trick:

1) If Hypotenuse is an odd number then follow the following
method.

For Example: Let us have number “9”
Step#1: Square it first of all i.e 92=81
Step#2: Now divide 81/2 which is 40.5

Step#3: Now add and subtract 0.5 to 40.5 i.e:
 40.5-0.5=40
 40.5+0.5=41
So, the triplets are 9,40, 41
2) If hypotenuse is an even number then follow the following

method:
For Example: Let us have number “8”
Step#1: Divide the number by 2 i.e 8/2 =4
Step#2:Now square it. i.e; 42=16
Step#3: Now add and subtract 1 from it. i.e;
 16-1=15
 16+1=17
So, the triplets are 4,15,17.

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Graph of Trigonometric Functions:

-Domain and Range of Trigonometric function:

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-Domain & Range of Inverse trigonometric Function:-

TRICK FOR GUESSING THE RELATIONS
OF EXERCISE 12.8

There are so many relations in exercise 12.8 and it is not
possible to memorize all of them for the mcq’s. But with
the following trick it is very easy to guess the relation.

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If a relation comes in mcq and you have to guess, then
first assume that the relation is for equilateral triangle.

Cram the following values.
1. α + β + r = 60° (As it is equilateral triangle)
2. s =
3. Δ = √ a²
4. r = √
5. R = √
6. r1 = r2 = r3 = √ a

Now, whatever the relation is, you can easily answer.

Example:-

r:R:r1=?
Put the values from above
√ :√ :√ a
Multiply by 2√3

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a : 2a : 3a

a will be cancelled

1:2:3

So, r : R : r1 = 1 : 2 : 3
Practice this on all questions of 12.8
Important Points:

-Important Results:
1) When two sides and the angle opposite to one of them are given;
In this case, either no triangle or one triangle or two triangles are
possible. For this reason, it is called ambiguous case.
Let b, c and B are given:
i) When B is acute and b<c sinB , no triangle is formed.
ii) When B is acute and b=c sinB, then only one triangle is formed which

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is right triangle.
iii) When B is acute and b>c sin B, then two triangles are formed if b<c
and only one triangle is possible iff b≥c.
iv) When B is obtuse, there is no triangle iff b<c and only one triangle iff
b>c.
Remarks:
1) The mid point of the hypotenuse of the right angle triangle is
equidistant from the three vertices of triangle.
2) In a right angle triangle, the orthocenter coincides with vertex
containing the right angle.
3) In a right triangle, the mid point of the hypotenuse is the circum
center of the triangle.
4) Point of congruency of three altitude is orthocenter of triangle.
5) Point of congruency of three median is called centroid triangle.
Important points:
1) ∆= ( − )( − )( − )
2) 2S=a+b+c
3) r:R:r1:r2:r3=1:2:3:3:3 (For Equilateral Triangle)
4) r1r2+r2r3+r3r1= s2
5) rr1r2r3=∆2
6) r1+r2+r3-r=4R
7) r1r2r3=rs2
8) r=R/2 (For equilateral triangle)
9) cosA/a= cosB/b =cosC/c (forms an equilateral triangle)
How to convert the radian into degree.
Example: 2π/3 radian into degree?
Solution: As we know that 1 rad= 180/π
so: 2π/3 * 180/π = 120*

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How to convert degree into radian:
Example: 54*45’ into radian
Solution:
As we know that 1*=60’
so (54 . 45/60)*= (54.3/4)*= (219/4)*
As we know that 1*=π/180
so; (219/4) (π/180)
=219π/720
= 0.955 (Answer)

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Properties of Triangle:-

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Some Trigonometric Tricks:

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Trick to find the Numerical value of trigonometric series:
Q#1:
sin1*. Sin2*. Sin3*…….. sin270=?
A) -1
B) 0 (Correct)
C) 1
D) None of these
Solution:
As we know that:
sin180*=0
The value of expression is 0. (Answer)
Q#2:
sin21*+ sin22*+…….+ sin289=?
Solution:
Trick: (Last angle/1st angle)= Product/2 = Answer
In this case;
89/1= 89/2= 44 ½ (Answer)
Q#3:
cos25+cos210+….. cos290=?
Solution:
Using the trick: (Last angle/1st angle)= Product/2= Answer.
In this case:
As cos290=0
so: cos25+…..+cos285=?
Now; 85/5= 17/2= 8 ½ (Answer)
Q#4:
The value of tan5*. Tan10*….tan85*=?

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Solution:
As we know that:
tanα=cotβ when α+β=90*
so that:
tan5=cot85
tan10= cot80
….
so that tan45 is one which can’t be paired with anyone so that
tan45*=1 (Answer)..
Q#5:
sinπ/9 . sin2π/9 .sinπ/3 .sin4π/9=?
A) ¾ B) 3/5 C) 3/17 D) 3/16 ( Correct)
Solution:
~Trick:
(Only, if one value is known)
Step#1: (Trigonometric value of angle)= P
Step#2: Take the square of its numerator.
Step#3: Take the power of denominator to the number of angles
involve in the notation.
Using the trick:
S#1: sin60= √

S#2: Squaring the numerator i.e (√3)2
S#3: Take the denominator power as the number of angles in notation
are 4 so: (2)4=16
Final result is 3/16 (Answer)
Q#6:
cos20.cos40. cos60.cos80=?
Solution:

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As we know that cos60=1/2
so using the trick:
= (1)2/24 = 1/16 (Answer)
Q#7: sin10.sin30.sin50.sin70=?
Solution: As we know that: sin30=1/2
So using the trick:
= (1)2/24=1/16 (Answer)
Q#8:
cos20.cos40.cos80=?
Solution:
There is a type where the exact angle is not mentioned in the question
so in that case we make it by adding any two angles whose sum is
known value. If the sum is from 30, 60, 90 etc then take the 1/2n for
final value whereas n=no. of angles involve in notation.
In this case; 20+40=60 so ½3=1/8 (Answer)
Q#9:
cosπ/5.cos2π/5.cos3π/5. Cos4π/5=?
Solution:
As 2π/5+ 3π/5= 5π/5= π
As we know that no. of angles in the notation are 4 so 1/24=1/16
(Answer).

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Some Basic Trigonometric Formulas:

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