SSC CGL Trigonometry Previous Year Questions PDF

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SSC CGL Trigonometry Previous Year Questions PDF

Download SSC CGL Trigonometry Questions with answers PDF based on previous papers very useful for SSC CGL exams. Very important Trigonometry Questions for SSC exams.

Question 1: A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height h. At a point on the plane, the angle of elevation of the bottom of the flag staff is $\alpha$ and that of the top of the flag staff is $\beta$. Then the height of the tower is

a) $h \tan \alpha$

b) $\frac{h \tan \alpha}{\tan \beta – \tan \alpha}$

c) $\frac{h \tan \alpha}{\tan \alpha – \tan \beta}$

d) None of these

Question 2: If $2y \cos \theta = x \sin \theta and 2x \sec \theta – y \cosec \theta = 3$, then the value of $x^2 + 4y^2$ is

a) 1

b) 2

c) 3

d) 4

Question 3: If $(a^2 – b^2)\sin \theta + 2ab \cos \theta = a^2 + b^2$, then $\tan \theta =$

a) $\frac{2ab}{a^2 – b^2}$

b) $\frac{a^2 – b^2}{2ab}$

c) $\frac{ab}{a^2 – b^2}$

d) $\frac{a^2 – b^2}{ab}$

Question 4: For $0^\circ < \theta < 90^\circ$, $tan\theta + cot\theta$ = 2 is equal to:

a) $30^\circ$

b) $60^\circ$

c) $45^\circ$

d) $0^\circ$

Question 5: If $\theta = 9^\circ$, then what is the value of
$\cot \theta \cot 2\theta \cot 3\theta \cot 4\theta \cot 5\theta \cot 6\theta \cot 7\theta \cot 8\theta \cot 9\theta$ ?

a) $\sqrt3$

b) $\sqrt3 – 1$

c) 1

d) $\frac{1}{\sqrt3}$

Question 6: $\theta$ being an acute angle, it is given that $\sec^2 \theta + 4 \tan^2 \theta = 6$. What is the value of $\theta$ ?

a) $60^\circ$

b) $45^\circ$

c) $0^\circ$

d) $30^\circ$

Question 7: What is the simplified value of $\frac{\sin^3 21^\circ + \cos^3 19^\circ}{\sin 21^\circ + \cos 19^\circ} + \sin^2 69^\circ + \cos^2 71^\circ + \frac{1}{\sec 69^\circ \cosec 71^\circ}$ is:

a) 2

b) 1

c) 4

d) 3

Question 8: If $\sin \theta + \cosec \theta = 2$, then what is the value of $\sin^{153} \theta + \cosec^{253} \theta$ ?

a) $\frac{1}{153 \times 253}$

b) $\frac{253}{153}$

c) 2

d) $\frac{153}{253}$

Question 9: If $\cos x = \frac{-\sqrt3}{2} and \pi < x < \frac{3\pi}{2}$, then the value of $4 \cot^2 x – 3 \cosec^2 x$ is:

a) 8

b) 0

c) 2

d) 1

Question 10: If $7(\cosec^2 55^\circ – \tan^2 35^\circ) + 2 \sin 90^\circ – \tan^2 52^\circ y \tan^2 38^\circ = \frac{y}{2}$, then the value of $y$ is:

a) 2

b) 3

c) 6

d) 1

$let \theta = 45\degree$

$2y \frac{1}{\sqrt{2}} = x\frac{1}{\sqrt{2}}$ = 2y= x

$2x \sqrt{2} – y \sqrt{2} = 3$

2x – y = $\frac{3}{\sqrt{2}}$ {substituting y= $\frac{x}{2}$}

x= $\sqrt{2}$

y= $\frac{1}{ \sqrt{2}}$

value of $x^2 + 4y^2$ = $\sqrt{2}^2 + 4\frac{1}{ \sqrt{2}^2}$ = 2+2 = 4

$(a^2 – b^2)\sin \theta + 2ab \cos \theta = a^2 + b^2$

divide it by $a^2 + b^2$

we get

$\frac{(a^2 – b^2)\sin \theta}{a^2 + b^2} + \frac{2ab \cos \theta}{a^2 + b^2} = 1$ ($\because \sin^2 \theta + \cos^2 \theta = 1$)

here $\sin \theta = \frac{(a^2 – b^2)}{a^2 + b^2}$

$\cos \theta = \frac{2ab}{a^2 + b^2}$

$\tan \theta = \frac{\sin \theta}{\cos \theta}$= $\frac{(a^2 – b^2)}{2ab}$

If $tan\theta + cot\theta$ = 2

Then $\tan\theta+\frac{1}{\tan\theta\ }=2$

If the sum of a number and its reciprocal is equal to then the number is equal to 1

So, $\tan\theta\ =1=45^{^{\circ\ }}$

Give, $\theta = 9^\circ$

$\cot \theta \cot 2\theta \cot 3\theta \cot 4\theta \cot 5\theta \cot 6\theta \cot 7\theta \cot 8\theta \cot 9\theta$

$\cot 9^\circ \cot 18^\circ \cot 27^\circ \cot 36^\circ \cot 45^\circ \cot54^\circ \cot63^\circ \cot 72^\circ \cot 81^\circ$

$\cot 9^\circ \cot 81^\circ\cot 18^\circ\cot 72^\circ\cot 27^\circ\cot63^\circ\cot 36^\circ \cot54^\circ\cot 45^\circ$

$\cot(90^\circ – 81^\circ)\cot81^\circ\cot(90^\circ – 72^\circ)\cot72^\circ\cot(90^\circ – 63^\circ)\cot63^\circ\cot(90^\circ – 54^\circ)\cot54^\circ\cot45^\circ$

$\tan81^\circ\cot81^\circ\tan72^\circ\cot72^\circ\tan63^\circ\cot63^\circ\tan54^\circ\cot54^\circ\cot45^\circ=1$.

As $\tan\theta\times\cot\theta=1 and \cot45^\circ=1$

So we get our answer as 1

$\sec^2 \theta + 4 \tan^2 \theta = 6$

$(1+tan^2 \theta) + 4 \tan^2 \theta = 6$

$5 \tan^2 \theta = 5$

$\tan^2 \theta=1$

$\tan \theta=1$

$\tan\theta=45^\circ$

$\theta=45^\circ$

$\sin \theta + \cosec \theta = 2$

$\sin \theta +\frac{1}{\sin \theta} = 2$

Let $\sin \theta = x$

$x+\frac{1}{x} = 2$

If sum of a number and its reciprocal is 2 then number will be 1

So, $\sin \theta = 1$

$\sin^{153} \theta + \cosec^{253} \theta$ = $1^{153}+1^{253}=2$

As Given in Question :

$\cos x = \frac {-\sqrt3}{2}$

$\Rightarrow \cos x = -\cos 30^{\circ}$

$\Rightarrow \cos x = -\cos 30^{\circ} = \cos (180 + 30)^{\circ}$     [ $\because -\cos \theta = \cos (180 + \theta ) ]$

$\therefore \cos x = \cos 210^{\circ}$

$\Rightarrow x=210^{\circ}$

Now  $4 \cot^2 x – 3 \cosec^2 x$

$\Rightarrow 4 \cot^2 210^{\circ} – 3 \cosec^2 210^{\circ}$

$\Rightarrow 4 (\sqrt3)^2 – 3 (-2)^2$  $[ \because \cot (180 + \theta) = \cot \theta , \cosec (180+\theta) = -\cosec \theta]$

$\Rightarrow 4\times3 – 3\times4$

$\Rightarrow 12 – 12 = 0$