# QUESTIONS ON TRIGONOMETRY (PART-3)

#### QUESTIONS ON TRIGONOMETRY (PART-3)

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 21

If Sin A = 3Sin (A+2B), then find the value of tan (A+B) + 2tan B

A) 1
B) 2
C) 4
D) 0

SHIV KISHOR #### QUERY 22

If sec@ + tan@ = 2+√5; then sin@ + cos@ = ?

A) √3
B) 1/√5
C) 3/√5
D) 2/√5

Amit Kumar
By squaring the given equation
sec²@ + tan²@ + 2sec@tan@ = 4 + 5 + 4√5

=> (1+tan²@) + tan²@ + 2sec@tan@ = 9 + 4√5

=> 2tan²@ + 2sec@tan@ = 8 + 4√5

=> 2tan@(tan@ + sec@) = 8 + 4√5

=> 2tan@(2+√5) = 4(2+√5) (given: sec@ + tan@ = 2+√5)

=> tan@ = 2

=> sin@ = 2cos@

We see that 3/√5 satisfies this condition when sin@ = 2/√5 and cos@ = 1/√5

Hence sin@ + cos@ = 3/√5 (option ‘C’)

#### QUERY 23

sina = 3sin(a+2b); then tan(a+b) + 2tanb?

A) 0
B) 1
C) 2
D) 3

SHIV KISHOR
Given that
sina = 3sin(a+2b)

=> sin(a+2b)/sina = 1/3                                     (dividing both sides by 3*sina)

=> [sin(a+2b) + sina]/[sin(a+2b) – sina] = [1+3]/[1-3] (by componendo and dividendo)

=> [2sin(a+b)cosb]/[2cos(a+b)sinb] = -2

=> tan(a+b)/tanb = -2

=> tan(a+b) + 2tanb = 0 (option ‘A’)

#### QUERY 24

If sec@ + tan@ = √3; then tan3@ is

A) undefined
B) 1/√3
C) considered as infinity
D) √3

SHIV KISHOR
sec @ + tan @ = √3
=> sec @ = √3 – tan @
Squaring
=> sec² @ = (√3 – tan² @
=> 1+ tan² @ = 3 + tan² @ – 2√3tan @
=> tan @ = 1/√3
=> @ = 30

Thus tan 3@ = tan 90°; which is infinity (option ‘C’)

#### QUERY 25

sin+ cosθ = √2 cosθ. Find value of cosθ – sinθ

A) √3 cosθ
B) √3 sinθ
C) √2 cosθ
D) √2 sinθ

SHIV KISHOR
sinθ + cosθ = √2cosθ
squaring
1 + sin2θ = 2cos²θ
= 2 – 2sin²θ
=2sin²θ = 1 – sin2θ
Thus, 2sin²θ = [cosθ – sinθ]²

Hence,
cosθ – sinθ = √2sinθ (option ‘D’)

#### QUERY 26

(sin 20)(sin 40)(sin 60)(sin 80) = ?

A) 1/16
B) 1/8
C) 3/16
D) 1/2

SHIV KISHOR
(sin 20)(sin 40)(sin 60)(sin 80)
=(√3/2)(1/2)(2sin 20)(sin 40)(sin 80)
= (√3/4)(cos 20 – cos 60)(sin80)
= (√3/8)[(2cos 20)(sin 80 – sin 80]
= (√3/8)[sin 100 – sin 80 + sin 60]
= (√3/8)[2sin 10)(cos 90 + √3/2]
= (√3/8)( 0 + √3/2)
= 3/16 (option ‘C’)

Another method
(sin A)(sin 2A)(sin 4A) = (1/4)(sin 3A)
Therefore, (sin 20)(sin 40)(sin 60)(sin 80)
= (sin 20)(sin 40)(sin 80)(sin 60)
= [(1/4)(sin 60)](sin 60)
= (1/4)((√3/2)(√3/2)
= 3/16 (option ‘C’)

#### QUERY 27

tan x = 4/3; then, sin x – cos x = ?

A) 1/5
B) 4/5
C) 3/5
D) 2/3

MAHA GUPTA
tan x = perpendicular/base
Therefore, the perpendicular = 4, and base = 3

By Pythagoras; hypotenuse = √(4² + 3²) = 5

Now sin x = perpendicular/hypotenuse = 4/5
And cos x = base/hypotenuse = 3/5

Hence, sin x – cos x = 4/5 – 3/5 = 1/5 (option ‘A’))

#### QUERY 28

If tanA ÷ cotA = 2, then the value of tan7A + cot7A is?

A) 2
B) 64
C) 16
D) 128

MAHA GUPTA
tanA/cotA = 2 (given)
=> tanA/(1/tanA) = 2
=> tanA = 1; as any expression in form of x+1/x = 2 will hold true only when x = 1, no matter what will be the power of x.

Therefore, tan7A + cot7A = tan7A + 1/tan7A = 2 (option ‘A’)

#### QUERY 29

If (sin A)/x = (cos A)/y; then sin A – cos A is?

A) x – y
B) x + y
C) (x – y)/(x² + y²)1/2
D) (y – x)/(x² + y²)1/2

MAHA GUPTA
(sin A)/x = (cos A)/y
=> sin A/cos A = x/y
=> tan A =x/y`
But tan A = perpendicular/base; means x is the perpendicular, and y the base

Therefore hypotenuse = √(x² + y²) i.e. (x² + y²)½

Now, sin A = perpendicular/hypotenuse = x/(x² + y²)½

And cos A = y/(x²+y²)½

Hence sin A – cos A = [x/(x² + y²)½] – [y/(x²)+(y²)½]
= (x – y)/(x²+y²)½ (option ‘C’)

#### QUERY 30

If ∠B of ΔABC is equal to 45°; find the value of (1+cot A)(1+cot C)

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

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