Question

If $\sin x=\frac{1}{4}$ and $x$ is in quadrant II, then
I. $\sin \frac{x}{2}=\frac{\sqrt{15}}{2}$
II. $\cos \frac{x}{2}=\frac{\sqrt{8-2 \sqrt{15}}}{4}$
III. $\tan \frac{x}{2}=4 \pm \sqrt{15}$

• A. I is correct
• B. II is correct
• C. III is correct
• D. All are correct

Answer 1

Answer: (B)

$\sin x=\frac{1}{4}$
Given that, $x$ is in second quadrant.
i.e., $\frac{\pi}{2}< x< \pi$
$ \because \sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}} $
$\therefore \frac{1}{4}=\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}$
$ \Rightarrow 1+\tan ^2 \frac{x}{2}=8 \tan \frac{x}{2}$
$ \Rightarrow \tan ^2 \frac{x}{2}-8 \tan \frac{x}{2}+1=0$
$ \Rightarrow \tan \frac{x}{2}=\frac{8 \pm \sqrt{64-4}}{2} $
$ \Rightarrow \tan \frac{x}{2}=\frac{8 \pm 2 \sqrt{15}}{2} \Rightarrow \tan \frac{x}{2}=4 \pm \sqrt{15}$
$\because \frac{\pi}{2} < x < \pi \Rightarrow \frac{\pi}{4}<\frac{x}{2}<\frac{\pi}{2}$
i.e., $\frac{x}{2}$ lies in Ist quadrant and as
$\frac{x}{2} \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right), \tan \frac{x}{2}>1 $
$\Rightarrow \tan \frac{x}{2}=4+\sqrt{15}$
Now, $ \sec ^2 \frac{x}{2}=1+\tan ^2 \frac{x}{2}=1+(4+\sqrt{15})^2$
$=1+16+15+8 \sqrt{15}$
$\Rightarrow \sec ^2 \frac{x}{2}=32+8 \sqrt{15}$
$\Rightarrow \sec ^2 \frac{x}{2}=8(4+\sqrt{15})$
$\Rightarrow \cos ^2 \frac{x}{2}=\frac{1}{\sec ^2 \frac{x}{2}}$
$=\frac{1}{8(4+\sqrt{15})} \times \frac{(4-\sqrt{15})}{(4-\sqrt{15})}$
$\Rightarrow \cos ^2 \frac{x}{2}=\frac{(4-\sqrt{15})}{8}$
$\sin x=\frac{1}{4}, \cos x=\frac{-\sqrt{15}}{4}$
$2 \sin ^2 \frac{x}{2}=1+\frac{\sqrt{15}}{4}=\frac{4+\sqrt{15}}{4}$
$\sin ^2 \frac{x}{2}=\frac{4+\sqrt{15}}{8}$
$\Rightarrow \sin \frac{x}{2}=\sqrt{\frac{4+\sqrt{15}}{8}}$