MATHSTRIGONOMETRIC RATIOS & IDENTITIESTRIGONOMETRY

QUESTIONS ON TRIGONOMETRY (PART-2)

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QUESTIONS ON TRIGONOMETRY (PART-2)

Most of these questions are taken from the previous year’s examinations conducted by the Staff Selection Commission (SSC) on Trigonometry (Trigonometric Ratios and Identities) of the Quantitative Aptitude (Maths) section of the following exams as well as other similar exams. They are all solved and supported by detailed explanation.

1. Combined Graduate Level (CGL) Exam Tier-I & Tier-II

2. Combined Higher Secondary (10+2) Exam (CHSL) Tier-I

3. SI in Delhi Police and CPO Exam Paper-I & Paper-II

QUERY 11

If sin 7x = cos 11x, then the value of tan 9x + cot 9x is?

A) 4
B) 1
C) 2
D) 3

Shivam Bajpai
sin 7x = cos (90 – 7x)
So, cos (90 – 7x) = cos 11x
=> 90 – 7x = 11x
=> x = 5

Now putting this in tan 9x + cot 9x
tan 45 + cot 45
= 1 + 1 = 2 (option ‘C’)


QUERY 12

In a triangle if sin² A + sin² B + sin² C = 2, then find whether the triangle is?

A) isosceles
B) right
C) acute
D) obtuse

JAYANTCHARAN CHARAN
The given equation is:
sin² A + sin² B + sin² C = 2

=> (1 – cos 2A)/2 + (1 – cos 2B)/2 + (1 – cos 2C)/2 = 2

=> 1 – cos 2A + 1 – cos 2B + 1 – cos 2C = 4

=> -1 = (cos 2A + cos 2B) + cos 2C

=> -1 = 2 cos (A + B)*cos (A – B) + cos 2(A + B)                                       [2C = 360 – 2(A+B)]

=> -1 = 2 cos (A + B)*cos (A – B) + 2 cos² (A + B) -1

=> 0 = 2 cos (A + B)*cos (A – B) + 2 cos² (A + B)

=> 0 = 2 cos (A+B) [cos (A – B) + cos (A + B)]

=> 0 = cos (A + B)*cos A*cos B

Means at least one of the three terms in the right hand side should be equal to 0, so either A + B = 90 or A = 90, or B = 90
Therefore it’s a right triangle. (option ‘B)


QUERY 13

If sin A + sin² A + sin³ A = 1; then find  cos6 A – 4cos4+ 8cos² A

A) 4
B) 3
C) 2
D) 1

RONNIE BANSAL
Given sin A + sin² A + sin³ A = 1

=> sin A + sin² A + sin³ A = cos² A + sin² A ……..(1 = cos² A + sin² A)

=> sin A + sin³ A = cos² A

=> sin A(1 + sin² A) = cos² A

=> sin A(1 + 1 – cos² A) = cos² A

Squaring both sides

sin² A(2 – cos² A)² = cos4 A

=> (1 – cos² A)( 4 + cos4 A – 4 cos² A) = cos4 A

=> 4 + cos4 A – 4 cos² A – 4cos² A – cos6 A + 4cos4 A= cos4 A

=> 4 = cos6 A – 4cos4 A – cos4 A + cos4 A + 4 cos² A + 4 cos² A

=> cos6 A – 4 cos4 A + 8 cos² A = 4 (option ‘A)


QUERY 14

[tan/(1 – cot)] + [cot/(1 – tan)] = ?

A) 1 – tan + cot
B) 1 + tan + cot
C) 1 – tan – cot
D) 1 + tan – cot

YOGESH DAHIYA
Convert all cot into tan
[tan/(1 – cot)] + [cot/(1 – tan)] = [tan/(1 – 1/tan)] + [(1/tan)/(1 – tan)]

= [tan²/(tan – 1)] + 1/tan(1 – tan)

= [tan²/(tan – 1)] – 1/tan(tan – 1)

= (tan³ – 1)/ tan(tan – 1)
= (tan³ – 1³)/tan(tan – 1)

= (tan – 1) (tan² + tan + 1)/tan(tan – 1) ————-(using a³ – b³ = (a – b)(a² + ab + b²)

Now cancelling tan – 1 in numerator and denominator
cot (tan² + tan + 1)

= tan + 1 + cot

= 1 + tan + cot (option ‘B’)


QUERY 15

If sin (x+y) = cos 3(x+y); then the value of tan 2(x+y) is ?

A) 1
B) 0
C) √(1/3)
D) √3

VAMSHI
Given sin (x+y) = cos 3(x+y)

=> Sin (x+y) = sin [90 – 3(x+y)]                           [Using [email protected] = sin (90 – @)]

=> x+y = 90 – 3x – 3y

=> 4(x+y) = 90

=> x+y = 90/4

Now substitute the value of x+y in tan 2(x+y)
tan 2(90/4) = tan 45 = 1 (option ‘A’)


QUERY 16

a sinQ + b cosQ = c; find a cosQ – b sinQ

A) √(a² + b² – c²)
B) √(a² – b² – c²)
C) (a² + b² + c²)
D) (a² + b² – c²)

MAHA GUPTA
a sinQ + b cosQ = c (given) ————–(1)
Let the find expression a cosQ – b sinQ = d ————–(2)

squaring & adding (1) & (2)
c² + d² = (a sinQ + b cosQ)² + (a cosQ – b sinQ)²

=> c² + d² = a²sin² Q + b²cos² Q + 2ab sinQ cosQ + a²cos² Q + b²sin² Q – 2ab sinQ cosQ

=> c² + d² = a²(sin² Q + cos² Q) + b²(sin² Q + cos² Q)

=> c² + d² = a²(1) + b²(1) [as we know sin² A + cos² A = 1]

=> c² + d² = a² + b²

=> d = √(a² + b² – c²)

But d = a cosQ – b sinQ (assumed)

So a cosQ – b sinQ= √(a² + b² – c²) (option ‘D’)


QUERY 17

Given 15 cot A = 8, find sin A and sec A

A) 14/17,  13/8
B) 15/17,  17/8
C) 17/8,  15/17
D) 17/8,  14/17

MAHA GUPTA
15 cot A = 8

=> cot A = 8/15

Means side adjacent to ∠A = 8; side opposite to ∠A = 15
So the hypotenuse = √(8² + 15²)—————–By Pythagoras theorem

=> 17

sine A = (side opposite to ∠A)/hypotenuse = 15/17 (option ‘B’)
sec A = hypotenuse/(side adjacent to ∠A) = 17/8 (option ‘B’)


QUERY 18

sin² 1 + sin² 89 = ?

A) o
B) 1
C) 2
D) 3

Bhushan Aggarwal
We know that: sin² A+cos² A=1.
Now sin² 1 + sin² 89
= sin² 1 + sin² (90 -1)
= sin² 1 + cos² 1 = 1 (option ‘B’)


QUERY 19

If tan α = n tan β and sin α = m sin β, then cos²α is?

1236944_626421447398515_1892365795_n

Neeraj Kumar Dhaneshri
sin α = m sin β                                             —- (i)
tan α = n tan β                                             —- (ii)

sin α/cos α = n sin β/cos β

by using (i)
m cos β = n cos α
=> n cos α = m cos β                                   —- (iii)

on squaring and adding (i) & (iii)
sin²α + n²cos²α = m²sin²β + m²cos²β
=> (n² – 1)cos²α = (m² – 1)
=> cos²α = (m² – 1)/(n² – 1)                      Option ‘b’

QUERY 20

The equation (cosP-1)x²+(cosP)x+sinP = 0, where x is a variable, has real roots. Then the interval of P may be one of the following?

A) 0, 2π
B) -π, 0
C) -π/2, π/2
D) 0, π

SHIV KISHOR
(cosP – 1) + (cosP)x + sinP = 0
If this equation has real roots then

(cosP)² ≥ 4(cosP – 1)sinP
or cos²P ≥ 4(cosP – 1)sinP

Since maximum and minimum values of cos²P are respectively +1 and 0, thus we are to find those values of ‘P’ for which
0 ≤ 4(cosP -1)sinP ≤ 1
which holds good only for 0 ≤ P ≤ π
Thus option ‘D’ is right.

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Maha Gupta

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