MATHSTRIGONOMETRIC RATIOS & IDENTITIESTRIGONOMETRY

QUESTIONS ON TRIGONOMETRY (PART-4)

QUESTIONS ON TRIGONOMETRY (PART-4)

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 31

If sec θ + tan θ = p, then (p² -1)(p² + 1) is equal to

A) sin θ
B) cos θ
C) sec θ
D) tan θ

SHIV KISHOR
sec@ + tan@ = p
=> 1 = (sec²@ + tan²@ + 2sec@tan@)1

applying componendo and dividendo rule
(p² -1)(p² + 1) = (sec²@ + tan²@ + 2sec@tan@ -1)(sec²@ + tan²@ + 2sec@tan@ +1)

= (2tan²@ + 2sec@tan@)(2sec²@ + 2sec@tan@)

= tan@sec@ = sin@ (option ‘A’)


QUERY 32

If 2 – cos²θ = 3sinθ*cosθ; sinθ is not equal to cosθ, then tanθ will be?

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

Aashish Gupta
2 – cos²θ = 3sinθcosθ

Divide both sides by cos²θ

2sec²θ -1 = 3tanθ

=> 2(1 + tan²θ) – 1 = 3tanθ

=> 2 + 2tan²θ – 1 = 3tanθ

=> 2tan²θ – 3tanθ + 1 = 0

=> 2tan²θ – 2tanθ –  tanθ + 1 = 0

=> 2tanθ(tanθ – 1) -1(tanθ -1) = 0

=> (2tanθ – 1)(tanθ – 1) = 0

=> tanθ = 1/2       or        tanθ = 1

But  tanθ = 1 is rejected as sinθ≠cosθ

Therefore tanθ = 1/2 (option ‘C’)

 
QUERY 33

In ΔABC, ∠A is a right angle and AD is perpendicular to BC. If AD = 4 cm, BC = 12 cm, then the value of (cotB + cotC) is?

A) 4
B) 3/2
C) 6
D) 3

SHIV KISHOR
1620724_562933480488351_8462567195255594356_n
cotB + cotC

= h4 + (12 – h)4
= 124 = 3 (option ‘D’)


QUERY 34

If 2sinX + 15cos²X = 7, then the value of cotX = ?

A) 3/4 or -√5/2
B) 3/4 or -2/√5
C) 4/3 or -√5/2
D) 4/3 or -2/√5

MAHA GUPTA
2sinX + 15cos²X + 7
=> 2sinX + 15(1 – sin²X) = 7
=> 2sinX = 7 – 15 + 15 sin²X
=> 15sin²X – 2sinX – 8 = 0
=> 15sin²X – 12 sinX + 10 sinX – 8 = 0
=> 3sinX(5sinX – 4) + 2(5sinX – 4) = 0
=> sinX = -2/3 or sinX = 4/5

  • For sinX = -2/3; hypotenuse = 3 and the sides are -2 and √5.
  • For sinX = 4/5, we can say that one side of the triangle is 4 and the hypotenuse is 5; so the other side must be 3.

Hence cosX = 35 or √53
Now cotX = cosX/sinX = (35)/(45) = 34 or [(√53)/(-23)] = -√52
So cotX = 34 or -√52 (option ‘A’)


QUERY 35

asec@ + btan@ + c = 0 and psec@ + qtan@ + r = 0;
then (br – qc)² – (pc – ar)² = ?

A) 1
B) 4
C) 0
D) 3

SHIV KISHOR
Given
asec@ + btan@ + c= 0 and
psec@ + qtan@ + r = 0
It’s possible only when a = p, b = q and c = r

Therefore
(br – qc)² – (pc – ar)²
= (bc – bc)² – (ac – ac)²
= 0 (option ‘C’)


QUERY 36

The numerical value of 1 + 1cot²63 – sec²27 + 1sin²63 – cosec²27

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

SHIV KISHOR
1 + 1cot²63 – sec²27 + 1sin²63 – cosec²27

= 1+ tan²63 – sec²27 + cosec²63 – cosec²27

= sec²63 – sec²27 + cosec²63 – cosec²27

=cosec²27 – cosec²63 + cosec²63 – cosec²27

= 0 (option ‘B’)


QUERY 37

If a cosQ + b sinQ = c, then (a sinQ – b cosQ)² = ?

A) a² + b² – c²
B) a² + b² + c²
C) a² – b² – c²
D) -a² + b² + c²

MAHA GUPTA
a cosQ + b sinQ = c (given)                                       —-(1)
Now, let a sinQ – b cosQ = d                                     —-(2)

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

=> c² + d² = a²cos²Q + b²sin²Q + 2ab sinQ cosQ + a²sin²Q + b²sin²Q – 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

=> d² = a² + b² – c²

But d = a sinQ – b cosQ              (assumed)

So (a sinQ – b cosQ)² = a² + b² – c² (option ‘A’)

QUERY 38

The area of a right triangle is 50. One of its angles is 45°. Find the lengths of the sides and hypotenuse of the triangle.

A) 10, 10√2
B) 10, 200
C) 15, 10√2
D) 10, 12

MAHA GUPTA
The triangle is right and the size one of its angles is 45°; means its third angle also has a size 45° and therefore the triangle is right and isosceles.

Now, let the length of one of the sides of the triangle = x
As the area of the triangle = 50
Therefore, 1× x² = 50
=> x = 10

Using Pythagoras, x² + x² = H²; where H is the hypotenuse of the triangle
10² + 10² = H²
=> H = 10√2

Hence, the length of the side = 10 and the length of the hypotenuse =  10√2 (option ‘A’)

QUERY 39

If tan θ = 2, then the value of 8sin θ + 5cos θsin³ θ + 2cos³ θ + 3cos θ is?

A) 215
B) 85
C) 75
D) 165

MAHA GUPTA
8sin θ + 5cos θsin³ θ + 2cos³ θ + 3cos θ

001 (2)

= sec² θ (8tan θ + 5)tan³ θ + 2 + 3sec² θ                                                  —- cos θcos³ θ = 1cos² θ = sec² θ

= 5(8*2 +5)2³ + 2 + 3*5                                                                           —- sec² θ = tan² θ +1 = 2² +1 = 5

= 21(option ‘A’)

QUERY 40

In a right triangle ABC with ∠A equal = 90°, find ∠B and ∠C so that sin(B) = cos(B).

A) ∠B = 30°, ∠C = 60°
B) ∠B = 40°, ∠C = 50°
C) ∠B = 45°, ∠C = 45°
D) ∠B = 60°, ∠C = 30°

MAHA GUPTA
Let b be the length of the side opposite angle B; c the length of the side opposite angle C; and h the length of the hypotenuse.

Then, sin(B) = b/h and cos(B) = c/h

As sin(B) = cos(B), means b/h = c/h
=> c = b
Means both the sides of the triangle are equal, means the triangle is isosceles
Therefore, both the angles B and C are 45° each (option ‘C’)

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Maha Gupta

Maha Gupta

Founder of www.examscomp.com and guiding aspirants on SSC exam affairs since 2010 when objective pattern of exams was introduced first in SSC. Also the author of the following books:

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