Question

If $\sin \alpha -\cos \alpha =\frac{1}{5}$,then $\tan \left(\frac{\alpha }{2}\right)$

Answer 1

$\sin \alpha -\cos \alpha =\frac{1}{5}$ given

$\Rightarrow \frac{2\tan \frac{\alpha }{2}}{1+\tan {}^2\frac{\alpha }{2}}-\frac{1-\tan {}^2\frac{\alpha }{2}}{1+\tan {}^2\frac{\alpha }{2}}=\frac{1}{5}\Rightarrow \frac{(2\tan \frac{\alpha }{2}+\tan {}^2\frac{\alpha }{2}-1)}{1+\tan {}^2\frac{\alpha }{2}}=\frac{1}{5}\Rightarrow 5\tan {}^2\frac{\alpha }{2}+10\tan \frac{\alpha }{2}-5=1+\tan {}^2\frac{\alpha }{2}\Rightarrow 4\tan {}^2\frac{\alpha }{2}+10\tan \frac{\alpha }{2}-6=0\Rightarrow 2\tan {}^2\frac{\alpha }{2}+5\tan \frac{\alpha }{2}-3=0\therefore \tan \frac{\alpha }{2}=\frac{-5\pm \sqrt{25+24}}{4}=\frac{4}{8},-3$

$\therefore \tan \frac{\alpha }{2}=\frac{1}{2}$ or $\tan \frac{\alpha }{2}=-3$