निम्नलिखित प्रत्येक प्रश्न में $\sin \frac{x}{2}, \cos \frac{x}{2}$ तथा $\tan \frac{x}{2},$ ज्ञात कीजिए

$\sin x=\frac{1}{4}, x$ द्वितीय चतुर्थांश में है

Here, $x$ is in quadrant $II$.

i.e., $\frac{\pi}{2} < x < \pi$

$\Rightarrow \frac{\pi}{4}<\frac{x}{2}<\frac{\pi}{2}$

Therefore, $\sin \frac{x}{2}, \cos \frac{x}{2}, \tan \frac{x}{2}$ are all positive.

It is given that $\sin x=\frac{1}{4}$

$\cos ^{2} x=1-\sin ^{2} x=1-\left(\frac{1}{4}\right)^{2}=1-\frac{1}{16}=\frac{15}{16}$

$\Rightarrow \cos x=-\frac{\sqrt{15}}{4}[\cos x \text { is negative in quadrant II }]$

$\sin ^{2} \frac{x}{2}=\frac{1-\cos x}{2}-\frac{1-\left(-\frac{\sqrt{15}}{4}\right)}{2}=\frac{4+\sqrt{15}}{8}$

$\Rightarrow \sin \frac{x}{2}=\sqrt{\frac{4+\sqrt{15}}{8}} \quad\left[\because \sin \frac{x}{2} \text { is negative }\right]$

$=\sqrt{\frac{4+\sqrt{15}}{8} \times \frac{2}{2}}$

$=\sqrt{\frac{8+2 \sqrt{15}}{16}}$

$=\frac{\sqrt{8+2 \sqrt{15}}}{4}$

$\cos ^{2} \frac{x}{2}=\frac{1+\cos x}{2}=\frac{1+\left(-\frac{\sqrt{15}}{4}\right)}{2}=\frac{4-\sqrt{15}}{8}$

$\Rightarrow \cos \frac{x}{2}=\sqrt{\frac{4-\sqrt{15}}{8}} \quad\left[\because \cos \frac{x}{2} \text { is positve }\right]$

$=\sqrt{\frac{4+\sqrt{15}}{8} \times \frac{2}{2}}$

$=\sqrt{\frac{8-2 \sqrt{15}}{16}}$

$=\frac{\sqrt{8-2 \sqrt{15}}}{4}$

$\tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\frac{\left(\frac{\sqrt{8+2 \sqrt{15}}}{4}\right)}{\frac{\sqrt{8-2 \sqrt{15}}}{4}}=\frac{\sqrt{8+2 \sqrt{15}}}{\sqrt{8-2 \sqrt{15}}}$

$=\sqrt{\frac{8+2 \sqrt{15}}{8-2 \sqrt{15}}} \times \frac{8+2 \sqrt{15}}{8+2 \sqrt{15}}$

$=\sqrt{\frac{(8+2 \sqrt{15})^{2}}{64-60}}=\frac{8+2 \sqrt{15}}{2}=4+\sqrt{15}$

Thus, the respective values are $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$

$\operatorname{are} \frac{\sqrt{8+2 \sqrt{15}}}{4}, \frac{\sqrt{8-2 \sqrt{15}}}{4}$ and $4+\sqrt{15}$