Question

If $sin\alpha +\cos \alpha =\frac{\sqrt{7}}{2},0<\alpha <\frac{\pi }{6},$ then $\tan \frac{\alpha }{2}$ is equal to

• A. $\frac{\sqrt{7}-2}{\sqrt{7}}$
• B. $\frac{1}{3}(\sqrt{7}-2)$
• C. $2-\sqrt{7}$
• D. $\frac{1}{3}(2-\sqrt{7})$

Answer 1

The correct option is B $\frac{1}{3}(\sqrt{7}-2)$

$\sin \alpha +\cos \alpha =\frac{\sqrt{7}}{2}$ ...... $\left(i\right)$
Now, $\sin 2\theta =\frac{2\tan \theta }{1+{\tan }^2\theta }$ and $\cos 2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }$
$\therefore \frac{2\tan \frac{\alpha }{2}}{1+{\tan }^2\frac{\alpha }{2}}+\frac{1-{\tan }^2\frac{\alpha }{2}}{1+{\tan }^2\frac{\alpha }{2}}=\frac{\sqrt{7}}{2}$
$2[2\tan \frac{\alpha }{2}+1-{\tan }^2\frac{\alpha }{2}]=\sqrt{7}[1+{\tan }^2\frac{\alpha }{2}]$
$4\tan \frac{\alpha }{2}+2-2{\tan }^2\frac{\alpha }{2}=\sqrt{7}+\sqrt{7}{\tan }^2\frac{\alpha }{2}$
${\tan }^2\frac{\alpha }{2}[\sqrt{7}+2]-4\tan \frac{\alpha }{2}+[\sqrt{7}-2]=0$
$\therefore \tan \frac{\alpha }{2}=\frac{4\pm \sqrt{16-4\left[3\right]}}{2[\sqrt{7}+2]}$
$\tan \frac{\alpha }{2}=\frac{4\pm 2}{2[\sqrt{7}+2]}=\frac{2\pm 1}{\sqrt{7}+2}$
$\tan \frac{\alpha }{2}=\frac{3}{\sqrt{7}+2}$ , $\tan \frac{\alpha }{2}=\frac{1}{\sqrt{7}+2}$ (on rationalising)
$\tan \frac{\alpha }{2}=\frac{3[\sqrt{7}-2]}{3}$ , $\tan \frac{\alpha }{2}=\frac{\sqrt{7}-2}{3}$
$\tan \frac{\alpha }{2}=\sqrt{7}-2$ , $\tan \frac{\alpha }{2}=\frac{\sqrt{7}-2}{3}$
$\sqrt{7}-2$ is rejected $\because 0<\alpha <\frac{\pi }{6}$
So, $0<\frac{\pi }{2}<\frac{\pi }{12}$
$\therefore \tan \left[\frac{\alpha }{2}\right]=\frac{1}{3}[\sqrt{7}-2]$