Question

Prove

$\frac{1+\sin \theta -\cos \theta }{1+\sin \theta +\cos \theta }=\tan \frac{\theta }{2}$

Answer 1

given, $\frac{1+\sin \theta -\cos \theta }{1+\sin \theta +\cos \theta }$

$=\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}+1-(1-2{\sin }^2\frac{\theta }{2})}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}+1+2{\cos }^2\frac{\theta }{2}-1}$

$=\frac{2\sin \frac{\theta }{2}(\cos \frac{\theta }{2}+\sin \frac{\theta }{2})}{2\cos \frac{\theta }{2}(\sin \frac{\theta }{2}+\cos \frac{\theta }{2})}$

$=\tan \frac{\theta }{2}$.