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.
Download Trigonometry Questions for RRB NTPC Set-3 PDF
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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
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Answers & Solutions:
1) Answer (A)
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
2) Answer (C)
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.
3) Answer (B)
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
$cos90=0$
4) Answer (D)
$(1 + \tan A) (1 + \tan B) = 2$
$ 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 $
5) Answer (A)
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6) Answer (A)
7) Answer (B)
$\frac{cot (90^\circ-\theta) sin(180^\circ-\theta) sec(360^\circ-\theta)}{tan(180^\circ+\theta) sec(-\theta) cos(90^\circ+\theta)}$
$\frac{cot (90^\circ-\theta)$ = tan(\theta)
$sin(180^\circ-\theta)$
$sec(360^\circ-\theta)$
$tan(180^\circ+\theta)$
$sec(-\theta)$
$cos(90^\circ+\theta)$
8) Answer (B)
In the second quadrant, sin $\theta$ varies from 1 to 0.
B is correct choice.
9) Answer (A)
$\sin\theta = \cos\theta when \theta=45\degree$
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)$\frac{-1}{2}$.
10) Answer (D)
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We hope this Top-10 Trigonometry Questions pdf for RRB NTPC exam will be highly useful for your Preparation.
