Trigonometry Questions for SSC CHSL Set-2 PDF

Download SSC CHSL Trigonometry  Questions with answers Set-2  PDF based on previous papers very useful for SSC CHSL Exams. Top-10 Very Important Questions for SSC Exam

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Question 1: In ΔABC, ∠C = 90° and AB = c, BC = a, CA = b; then the value of (cosec B – cos A) is

a) $\frac{c^2}{ab}$

b) $\frac{b^2}{ca}$

c) $\frac{a^2}{bc}$

d) $\frac{bc}{a^2}$

Question 2: If tanθ = 1/√11 0 < θ < π/2, then the value of $\frac{cose{c}^2\theta -{\sec }^2\theta }{cose{c}^2\theta +{\sec }^2\theta }$

a) $\frac{3}{4}$

b) $\frac{4}{5}$

c) $\frac{5}{6}$

d) $\frac{6}{7}$

Question 3: $\frac{sin\theta }{x}=\frac{cos\theta }{y}$, then $sin\theta -cos\theta$ is equal to

a) $x-y$

b) $x+y$

c) $\frac{x-y}{\sqrt{x^2+y^2}}$

d) $\frac{y-x}{\sqrt{x^2+y^2}}$

Question 4: If x sin 60°.tan 30° = sec 60°.cot 45°, then the value of x is

a) 2

b) 2√3

c) 4

d) 4√3

Question 5: A sphere and a hemisphere have the same radius. Then the ratio of their respective total surface areas is

a) 2 : 1

b) 1 : 2

c) 4 : 3

d) 3 : 4

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Question 6: If x*sin^3 θ + y *cos^3 θ = sinθ * cosθ ≠ 0 and x sinθ – y cosθ = 0, then value of (x^2 + y^2 ) is

a) 1

b) sinθ – cosθ

c) sinθ + cosθ

d) 0

Question 7: If 2 – cos^2 θ = 3 sin θ cos θ, sin θ ≠ cos θ then tan θ is

a) 1/2

b) 0

c) 2/3

d) 1/3

Question 8: If $tan\theta =\frac{a}{b}$, then $\frac{asin\theta +bcos\theta }{asin\theta –bcos\theta }$ is

a) $\frac{a}{a^2+b^2}$

b) $\frac{b}{a^2+b^2}$

c) $\frac{a^2-b^2}{a^2+b^2}$

d) $\frac{a^2+b^2}{a^2-b^2}$

Question 9: If $sin\theta +cos\theta =\sqrt{2}sin({90}^{\circ }–\theta )$, then cot θ is equal to

a) $\sqrt{2}+1$

b) $\frac{1}{\sqrt{2}}+2$

c) $\sqrt{20}-1$

d) None of these

Question 10: What is the value of cosec 11π/6?

a) -2/√3

b) √2

c) -2

d) -√2

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Answers & Solutions:

1) Answer (C)

In $\mathrm{△}$$ABC,A{B}^2=A{C}^2+B{C}^2$

=> $c^2=a^2+b^2=>c^2–b^2=a^2$

$cosecB=\frac{AB}{BC}=\frac{c}{b}$

$cosA=\frac{AC}{AB}=\frac{b}{c}$

$\therefore cosecB–cosA=\frac{c}{b}–\frac{b}{c}$

= $\frac{c^2-b^2}{bc}=\frac{a^2}{bc}$

2) Answer (C)

Expression : $tan\theta =\frac{1}{\sqrt{11}}$

We know that, $sec\theta =\sqrt{1+ta{n}^2\theta }$

=> $sec\theta =\sqrt{1+\frac{1}{11}}=\sqrt{\frac{12}{11}}$

Now, $cosec\theta =\frac{sec\theta }{tan\theta }$

=> $cosec\theta =\sqrt{12}$

To find : $\frac{cose{c}^2\theta -{\sec }^2\theta }{cose{c}^2\theta +{\sec }^2\theta }$

= $\frac{12–\frac{12}{11}}{12+\frac{12}{11}}$

= $\frac{1–\frac{1}{11}}{1+\frac{1}{11}}$

= $\frac{10}{12}=\frac{5}{6}$

3) Answer (C)

$\frac{sin\theta }{x}=\frac{cos\theta }{y}$

Reaaranging the given data , we get

$tan\theta$ = $\frac{x}{y}$

Now taking $cos\theta$ common from $sin\theta -cos\theta$,we get

= $cos\theta (tan\theta )–1$…………(1)

Imagine a right angle triangle

From this triangle , we can calculate values of $cos\theta$ and $tan\theta$ and hence putting the values in equation 1

we get = $\frac{y}{\surd (x^2+y^2)}$ ($\frac{x}{y}$- 1)

= $\frac{x-y}{\sqrt{x^2+y^2}}$

4) Answer (C)

sin 60° = $\frac{\sqrt{(}3)}{2}$

tan 30 = $\frac{1}{\sqrt{(}3)}$

sec 60 = 2

cot 45 = 1

x $\frac{\sqrt{(}3)}{2}$ $\frac{1}{\sqrt{(}3)}$ = $(2\times 1)$

x = 4

5) Answer (C)

Let the common radius be “r” units

area of sphere = 4$\pi r^2$

area of hemi-sphere = 2$\pi r^2$ + $\pi r^2$ = 3$\pi r^2$

$\frac{\text{Area of Sphere}}{\text{Area of Hemisphere}}$ = $\frac{4\pi r^2}{3\pi r^2}$

= $\frac{4}{3}$

6) Answer (A)

Given, xsinθ -ycosθ = 0
y=xsinθcosθ

⇒y=xsinθcosθ

Given the expression:
xsin${}^3$θ + ycos${}^3$θ = sinθcosθ
Replacing the value of y we get,
xsin${}^3$θ+(xsinθ/cosθ)×cos${}^3$θ=sinθcosθ
x sin3θ + x sinθcos${}^2$θ = sinθcosθ
x sinθ × (sin2θ + cos2θ) = sinθcosθ
We know the identity: sin2θ + cos2θ = 1
x = cosθ
y=xsinθ/cosθ
y = sinθ
Hence,
x${}^2$ + y${}^2$ = cos${}^2$θ + sin${}^2$θ
= 1
Option A is the correct answer.

7) Answer (A)

$2-co{s}^2\theta =1+1-co{s}^2\theta =1+si{n}^2\theta$
Dividing the LHS and RHS by $co{s}^2\theta$
$1+si{n}^2\theta =se{c}^2\theta +ta{n}^2\theta$
$3sin\theta cos\theta =3tan\theta$
$se{c}^2\theta +ta{n}^2\theta =3tan\theta$
$se{c}^2\theta =1+ta{n}^2\theta$
$1+ta{n}^2\theta +ta{n}^2\theta =3tan\theta$
$1+2ta{n}^2\theta =3tan\theta$
$2ta{n}^2\theta –3tan\theta +1=0$
let x=$tan\theta$
The equation becomes
$2x^2+3x+1=0$
On solving for x we get x=$1$ and x=$\frac{1}{2}$
Option A is the correct answer.

8) Answer (D)

To find : $\frac{asin\theta +bcos\theta }{asin\theta –bcos\theta }$

Dividing numerator and denominator by $cos\theta$, we get :

= $\frac{atan\theta +b}{atan\theta –b}$

Also, it is given that $tan\theta =\frac{a}{b}$

= $\frac{a\times \frac{a}{b}+b}{a\times \frac{a}{b}–b}$

= $\frac{\frac{a^2+b^2}{b}}{\frac{a^2–b^2}{b}}$

= $\frac{a^2+b^2}{a^2–b^2}$

9) Answer (A)

Expression : $sin\theta +cos\theta =\sqrt{2}sin({90}^{\circ }–\theta )$

=> $sin\theta +cos\theta =\sqrt{2}cos\theta$

=> $sin\theta =cos\theta (\sqrt{2}–1)$

=> $\frac{cos\theta }{sin\theta }=\frac{1}{\sqrt{2}–1}$

=> $cot\theta =\frac{1}{\sqrt{2}–1}\times \frac{\sqrt{2}+1}{\sqrt{2}+1}$

=> $cot\theta =\sqrt{2}+1$

10) Answer (C)

Expression : cosec 11π/6

= $cosec(2\pi –\frac{\pi }{6})$

= $-cosec(\frac{\pi }{6})=-2$

=> Ans – (C)

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