Trigonometry Questions for RRB NTPC Set-3 PDF

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Trigonometry Questions for RRB NTPC Set-3 PDF

Download RRB NTPC Top-10 Trigonometry Questions Set-3 PDF. Questions based on asked questions in previous exam papers very important for the Railway NTPC exam.

Question 1: If $\sec \theta + \tan \theta = 2,$then $\sec \theta – \tan \theta = ?$

a) 0.5

b) 1.5

c) 1

d) 0.75

Question 2: If $\sin A = \cos B$, where $A$ and $B$ are acute, then find $(A + B)$.

a) $60^\circ$

b) $50^\circ$

c) $90^\circ$

d) $45^\circ$

Question 3: If A and B are complementary angles, then find the value of $\cos A \cos B – \sin A \sin B$.

a) 1

b) 0

c) $\frac{1}{2}$

d) -1

Question 4: If $(1 + \tan A) (1 + \tan B) = 2$, then find $A + B$.

a) $90^\circ$

b) $180^\circ$

c) $0^\circ$

d) $45^\circ$

Question 5: If A=$30^\circ$ then find the value of $cos^{2}A+cos^{2}(60+A)+cos^{2}(60-A)$

a) $\frac{3}{2}$

b) $\frac{\sqrt{3}}{2}$

c) $\frac{-3}{2}$

d) 1

Question 6: If and b are supplementary angles find the value of
$\frac{\tan A + \tan B}{1 – \tan A \tan B}$

a) 0

b) 1

c) -1

d) $\frac{1}{2}$

Question 7: simplify $\frac{cot (90^\circ-\theta) sin(180^\circ-\theta) sec(360^\circ-\theta)}{tan(180^\circ+\theta) sec(-\theta) cos(90^\circ+\theta)}$

a) 2

b) -1

c) 1

d) -2

Question 8: In the second quadrant, sin $\theta$ varies from:

a) 1 to -1

b) 1 to 0

c) 0 to -1

d) -1 to 1

Question 9: $If \sin\theta = \cos\theta, and \theta$ is acute, the value of  $2\sin^{2} \theta – 3\cos^{2} \theta is :$

a) $-\frac{1}{2}$

b) 1

c) -1

d) $\frac{1}{2}$

Question 10: If $\cosec \theta + \cot \theta = 2,$ then $\cot \theta = ?$

a) 0

b) 0.5

c) 1

d) 0.75

The correct option is A.

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

$\Rightarrow$ ($sec\theta+ tan\theta$)($sec\theta – tan\theta$) = 1

$\Rightarrow$             2 $\times$ ( $sec\theta – tan\theta$) = 1

$\Rightarrow$.            $sec\theta – tan\theta$ = 0.5

Given, Sin A = Cos B, it can be written as

Sin A = sin(90 – B)

A = 90 – B

A + B = 90.

C is correct choice.

If A and B are complementary angles then

A+B = 90

$\cos A \cos B – \sin A \sin B$.

we know that

$cos(A+B)=\cos A \cos B – \sin A \sin B$.

cos(A+B)

put value A+B=90

$cos90=0$

$(1 + \tan A) (1 + \tan B) = 2$

$1 + \tan B + \tan A + \tan A \tan B = 2$

$\tan A + \tan B + \tan A \tan B = 1$

$\tan A + \tan B = 1 – \tan A \tan B$

$\frac{ \tan A + \tan B}{1 – \tan A \tan B} = 1$

$\frac{ \tan x + \tan y}{1 – \tan x \tan y} = \tan (x+y)$

$\frac{ \tan A + \tan B}{1 – \tan A \tan B} = \tan(A+B) = 1$

$\tan (A+B) = \tan 45^\circ$

therefore,$A+B = 45^\circ$

$\frac{cot (90^\circ-\theta) sin(180^\circ-\theta) sec(360^\circ-\theta)}{tan(180^\circ+\theta) sec(-\theta) cos(90^\circ+\theta)}$

$\frac{cot (90^\circ-\theta)$ = tan(\theta)

$sin(180^\circ-\theta)$

$sec(360^\circ-\theta)$

$tan(180^\circ+\theta)$

$sec(-\theta)$

$cos(90^\circ+\theta)$

In the second quadrant, sin $\theta$ varies from 1 to 0.

B is correct choice.

$\sin\theta = \cos\theta when \theta=45\degree$

So , $2\sin^{2} \theta – 3\cos^{2} \theta$

=$\frac{2}{2} – \frac{3}{2}$

=$\frac{-1}{2}$

So, the answer would be option b)$\frac{-1}{2}$.