RRB NTPC CBT 2 Level 5 June-12-2022 Shift 1 Exam Previous Question Paper with Solutions

16) If √3 tan 2θ – 3 = 0, then find the value of tanθ secθ-sinθ where 0<θ<90°

A) $\frac76$
B) $\frac56$
C) $\frac23$
D) $\frac16$

View Answer

D) $\frac16$

Explanation:Let’s solve the equation step by step.
Given equation:
$\sqrt{3} \tan(2\theta) - 3 = 0$
We need to find the value of $\tan\theta \sec\theta - \sin\theta$ .
Step 1: Solve for $\tan(2\theta)$
From the given equation:
$\sqrt{3} \tan(2\theta) = 3$
$\tan(2\theta) = \frac{3}{\sqrt{3}} = \sqrt{3}$
Now, recall that:
$\tan(2\theta) = \sqrt{3}$
This implies that $2\theta = 60^\circ$ because $\tan(60^\circ) = \sqrt{3}$ .
Step 2: Solve for $\theta$
From $2\theta = 60^\circ$ , we get:
$\theta = 30^\circ$
Step 3: Find $\tan\theta$ , $\sec\theta$ , and $\sin\theta$ for $\theta = 30^\circ$
$\tan(30^\circ) = \frac{1}{\sqrt{3}}$
$\sec(30^\circ) = \frac{2}{\sqrt{3}}$
$\sin(30^\circ) = \frac{1}{2}$
Step 4: Calculate $\tan\theta \sec\theta - \sin\theta$
Now, we need to calculate:
$\tan\theta \sec\theta - \sin\theta$
Substitute the values:
$\tan(30^\circ) = \frac{1}{\sqrt{3}}, \quad \sec(30^\circ) = \frac{2}{\sqrt{3}}, \quad \sin(30^\circ) = \frac{1}{2}$
$\left( \frac{1}{\sqrt{3}} \right) \left( \frac{2}{\sqrt{3}} \right) - \frac{1}{2}$
First, calculate $\frac{1}{\sqrt{3}} × \frac{2}{\sqrt{3}}$ :
$\frac{1}{\sqrt{3}} × \frac{2}{\sqrt{3}} = \frac{2}{3}$
So, the expression becomes:
$\frac{2}{3} - \frac{1}{2}$
Now, subtract $\frac{1}{2}$ from $\frac{2}{3}$ :
$\frac{2}{3} - \frac{1}{2} = \frac{4}{6} - \frac{3}{6} = \frac{1}{6}$
Final Answer:
The value of $\tan\theta \sec\theta - \sin\theta$ is $\frac{1}{6}$ .
Thus, the correct answer is:
Option $\frac{1}{6}$ .