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 $\frac{x}{2}<\frac{\pi }{8}$ ]
$=\frac{\sqrt{7}-2}{3}$