Question

Simplify the following expression:
$\tan (\frac{\alpha }{2}+\frac{\pi }{4})\frac{1-\sin \alpha }{\cos a}.$

Answer 1

$\tan (\frac{\alpha }{2}+\frac{\pi }{4})\left(\frac{1-\sin x}{\cos x}\right)$

$=\frac{\tan \frac{\alpha }{2}+\tan \frac{\pi }{4}}{1-\tan \frac{\pi }{4}\tan \frac{\alpha }{2}}\times \frac{{\sin }^2\frac{\alpha }{2}+{\cos }^2\frac{\alpha }{2}-2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}{{\cos }^2\frac{\alpha }{2}-{\sin }^2\frac{\alpha }{2}}$

$=\frac{1+\tan \frac{\alpha }{2}}{1-\tan \frac{\alpha }{2}}\times \frac{{(-\sin \frac{\alpha }{2}-\cos \frac{\alpha }{2})}^2}{(\cos \frac{\alpha }{2}-\sin \frac{\alpha }{2})(\cos \frac{\alpha }{2}+\sin \frac{\alpha }{2})}$

$=\frac{\sin \frac{\alpha }{2}+\cos \frac{\alpha }{2}}{\cos \frac{\alpha }{2}-\sin \frac{\alpha }{2}}\times \frac{\cos \frac{\alpha }{2}-\sin \frac{\alpha }{2}}{\sin \frac{\alpha }{2}+\cos \frac{\alpha }{2}}=1$