Question

If $\sin x+\cos x=\frac{\sqrt{7}}{2}$, where $x \in A$, then $\tan \frac{x}{2}$ is equal to

• A. $\frac{3-\sqrt{7}}{3}$
• B. $\frac{\sqrt{7}-2}{3}$
• C. $\frac{4-\sqrt{7}}{4}$
• D. none of these

Answer 1

Answer: (B)

$\sin x+\cos x=\frac{\sqrt{7}}{2}$
$\Rightarrow \frac{2 \tan \frac{x}{2}}{\left(1+\tan ^{2} \frac{x}{2}\right)}+\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}=\frac{\sqrt{7}}{2}$
or $(\sqrt{7}+2) \tan ^{2} \frac{x}{2}-4 \tan \frac{x}{2}+(\sqrt{7}-2)=0$
or $\tan \frac{x}{2} =\frac{4 \pm \sqrt{16-4(7-4)}}{2(\sqrt{7}+2)}$
$=\frac{1}{(\sqrt{7}+2)}$ [as $\frac{x}{2} < \frac{\pi}{8}$ ]
$=\frac{\sqrt{7}-2}{3}$