0
453

# 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

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 $\frac{cosec^{2}\theta-\sec^2\theta}{cosec^2\theta+\sec^2\theta}$

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

SSC CHSL Study Material (FREE Tests)

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 $tanθ = \frac{a}{b}$, then $\frac{a sin \theta + b cos \theta}{a sin \theta – b cos\theta}$ 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 $sin θ + cos θ = \sqrt{2} sin (90^{\circ} – θ)$, 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

In $\triangle$$ABC, AB^2 = AC^2 + BC^2$

=> $c^2 = a^2 + b^2 => c^2 – b^2 = a^2$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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)

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

sin 60° = $\frac{\sqrt(3)}{2}$

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

sec 60 = 2

cot 45 = 1

x $\frac{\sqrt(3)}{2}$ $\frac{1}{\sqrt(3)}$ = $(2\times1)$

x = 4

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$\pi r^2$

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

= $\frac{4}{3}$

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.

$2-cos^2 \theta = 1+1-cos^2 \theta = 1+sin^2 \theta$
Dividing the LHS and RHS by $cos^2\theta$
$1+sin^2 \theta = sec^2 \theta + tan^2 \theta$
$3sin \theta cos \theta = 3 tan \theta$
$sec ^2 \theta + tan^2 \theta = 3tan \theta$
$sec ^2 \theta = 1+tan^2 \theta$
$1+tan^2 \theta+tan^2 \theta = 3tan \theta$
$1+2tan^2 \theta = 3tan \theta$
$2tan^2 \theta – 3tan \theta + 1 =0$
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.

To find : $\frac{a sin \theta + b cos \theta}{a sin \theta – b cos\theta}$

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

= $\frac{a tan \theta + b}{a tan \theta – b}$

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

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

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

= $\frac{a^2 + b^2}{a^2 – b^2}$

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

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

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

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

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

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

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