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) 60^o
(B) 30^o
(C) 15^o
(D) 45^o

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Ans : (A) 60^o
$3 cot\theta + tan \theta -2\sqrt3 = 0$
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(48^o + 2x), and 0^o <x <90^o , then what is the value of x ?
(A) 14^o
(B 12^o
(C) 21^o
(D) 16^o

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Ans : (A) 14^o
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) 45^o
(B) 30^o
(C) 60^o
(D) 75^o

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

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) 30^o
(B) 15^o
(C) 60^o
(D) 45^o

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Ans : (C) 60^o
$sec^2\theta +(1 -\sqrt3) tan\theta -(1 +\sqrt3) =0$
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

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) $\frac{25}{12}$

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

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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) $\frac{\sqrt21}{3}$