Trigonometry Questions for Competitive Exams

Trigonometry Questions with solution for Competitive Exams. Previous year Exam Question with official answer and detail explanation with short tricks. The Questions are as per SSC CGL and CHSL Exam Syllabus and Pattern. The below 1-18 questions are from the all shifts of SSC CHSL Exam August 2021.

Trigonometry Questions with Solution

Q.1: If (sin A – cos A) = 0 , then what is the value of cot A ?
(A) \frac{\pi}{6}
(B) 0
(C) 1
(D) \frac{\pi}{4}

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Ans : (C) 1
(sin A – cos A) = 0
sin A = cos A
\frac{cos A}{sin A} = 1
cot A = 1

Q.2: If cosec\theta=\frac{41}{9} and \theta is an acute angle, then the value of 5 tan\theta will be:
(A) \frac 98
(B) \frac {11}{8}
(C) \frac {13}{4}
(D) \frac 78

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Ans : (A) \frac 98
cosec\theta=\frac{41}{9}
sin\theta=\frac{9}{41}
पाइथागोरस प्रमेय से
BC2 = 412 – 92
BC =\sqrt{1600} = 40
ATQ 5 tan\theta = 5\times \frac{9}{40}
=\frac98

Q.3: Solve the following equation and find the value of \theta.
3 cot\theta + tan \theta -2\sqrt3 = 0,0 <\theta < 90^o
(A) 60o
(B) 30o
(C) 15o
(D) 45o

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Ans : (A) 60o
3 cot\theta + tan \theta -2\sqrt3 = 0
Put \theta = 60 its satisfy
3 x cot60 + tan60 -2 \sqrt{3}
3\times\frac{1}{\sqrt3} +\sqrt3 - 2\sqrt3
\frac{3}{\sqrt3} +\sqrt3 - 2\sqrt3
\frac{6}{\sqrt3} -2\sqrt3
\frac {6-6}{\sqrt3} = \frac{0}{\sqrt3} = 0
so \theta = 60

Q.4: In \triangle ABC , \angle A = 90^0, AB = 20cm and BC = 29cm. What is the value of (sinB -cotC) ?
(A) \frac{189}{580}
(B) -\frac {9}{29}
(C) \frac{9}{29}
(D) -\frac{189}{580}

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Ans : (D) -\frac{189}{580}

Trigonometry Questions for Competitive Exams

Q.5: If tanx = cot(48o + 2x), and 0o <x <90o , then what is the value of x ?
(A) 14o
(B 12o
(C) 21o
(D) 16o

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Ans : (A) 14o
tan x = cot (48 + 2x)
tanx = tan (90 – (48 + 2x)
x = 90 -(48 + 2x)
x = 90 – 48 – 2x
3x = 42
x = 14

Q.6: If 7 sin^2\theta + 3cos^2\theta = 4,0^o <\theta <90^o , then the value of \theta will be:
(A) 45o
(B) 30o
(C) 60o
(D) 75o

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Ans : (B) 30o
7 sin^2\theta + 3cos^2\theta = 4
4 sin^2\theta + 3sin^2\theta +3cos^2\theta= 4
4 sin^2\theta + 3(sin^2 \theta + cos^2\theta)= 4
4sin^2\theta + 3\times1 = 4
4sin^2\theta = 4 - 3 = 1
sin^2\theta =\frac 14
sin\theta =\frac 12
\theta = 30^o

Q.7: Find the value of \theta , If sec2 sec^2\theta +(1 -\sqrt3)tan\theta -(1 +\sqrt3)=0, where \theta is an acute angle.
(A) 30o
(B) 15o
(C) 60o
(D) 45o

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Ans : (C) 60o
sec^2\theta +(1 -\sqrt3) tan\theta -(1 +\sqrt3) =0
put \theta = 60^o (By option)
sec^2(60) + (1 -\sqrt3) tan 60 - (1 +\sqrt3)
4 +(1 - \sqrt3) \sqrt3 - (1 + \sqrt3)
4 +\sqrt3 - 3 - 1 -\sqrt3
4 – 4 = 0
So \theta = 60

Q.8: If cos\theta =\frac{7}{3\sqrt6} and \theta is an acute angle, then the value of 27 sin^2\theta -\frac 32 is:
(A) 15
(B) 12
(C) 1
(D) 9

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Ans : (C) 1

Trigonometry Questions for Competitive Exams

Q.9: In a triangle ABC , right-angled at C, if sec A =\frac{13}{5} , then find the value of \frac{1+ sinA}{cosB}.
(A) \frac{18}{5}
(B) 5
(C) \frac32
(D) \frac{25}{12}

[toggle] Ans : (D) \frac{25}{12}

Trigonometry Questions for Competitive Exams

Q.10: If sin\theta + cosec \theta = 7 , then what is the value of sin^3\theta + cosec^3\theta ?
(A) 322
(B) 382
(C) 367
(D) 350

Show Answer
Ans : (A) 322
sin\theta + cosec \theta = 7
(sin\theta + cosec \theta)^3 = 7^3
sin^3\theta + cosec^3 \theta +3sin\theta cosec\theta\times7 = 343
sin^3\theta + cosec^3 \theta +3\times7 = 343 [Sin\theta\times Cosec\theta = 1]
sin^3\theta + cosec^3\theta = 343 - 21 = 322
Trigonometry Questions for Competitive Exams

Q.11: If 3 cot A = 4 tan A , and A is an acute angle, then what will be the value of Sec A ?
(A) \frac{\sqrt7}{2}
(B) \frac{1}{\sqrt3}
(C) \frac12
(D) \frac{\sqrt21}{3}

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Ans : (A) \frac{\sqrt7}{2}

Q.12: What is the value of \frac{cos^220^o + cos^270^o}{sin^290^o} -tan^245^o ?
(A) -1
(B) 0
(C) -2
(D) 1

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Ans : (B) 0
\frac{cos^220^o + cos^270^o}{sin^290^o} -tan^245^o
\frac{1}{1} -1 [Cos220 + Cos270 = 1, tan 45 = 1]
= 0

Q.13: If \sqrt{13} Sin \theta = 2 , then the value of \frac{3tan\theta +\sqrt{13}Sin\theta}{\sqrt{13}Cos \theta-3 tan \theta} is:
(A) 5
(B) \frac 12
(C) 3
(D) 4

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Ans : (D) 4

Q.14: In a right-angled triangle ABC right angled at C, Sin A = Sin B. What is the value of Cos A ?
(A) 1
(B) \frac{\sqrt3}{2}
(C) \frac 12
(D) \frac{1}{\sqrt2}

Show Answer
Ans : (D) \frac{1}{\sqrt2}
Sin A = Sin B
A = B = 45 Satisfy
Cos A = Cos 45
= \frac{1}{\sqrt2}

Q.15: If 5 Cos \theta = 4 Sin \theta , 0^o \leq \theta \leq 90^o , then what will be the value of Sec \theta ?
(A) \frac{\sqrt{41}}{4}
(B) \frac{\sqrt{41}}{16}
(C) \frac{\sqrt{41}}{5}
(D) \frac35

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Ans : (A) \frac{\sqrt{41}}{4}

Q.16: The value of \frac{Cos 8^o Cos 24^o Cos 60^o Cos 66^o Cos 82^o}{Sin 82^o Sin 66^o Sin 60^o Sin 8^o Sin 24^o} is:
(A) \frac{1}{\sqrt2}
(B) \frac{1}{\sqrt3}
(C) 1
(D) 0

Show Answer
Ans : (B) \frac{1}{\sqrt3}
Cos 8o Cos 82o = 1
Cos 24o Cos 60o = 1
Sin 82o Sin 8o = 1
Sin 66o Sin 24o = 1
so \frac{Cos 66^o}{Sin 60^o}
Cot 60=\frac{1}{\sqrt3}

Q.17: If Sin B =\frac{9}{41} , then what is the value of Cot B, where 0o <B<90o ?
(A) \frac{9}{41}
(B) \frac{41}{9}
(C) \frac{9}{40}
(D) \frac{40}{9}

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Ans : (D) \frac{40}{9}

Q.18: If tan\theta =\frac43 , then the value of \frac{9 Sin\theta + 12 Cos \theta}{27 Cos\theta -20sin \theta} will be equal to:
(A) 36
(B) 100
(C) 18
(D) 72

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Ans : (D) 72

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