Question

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

Answer 1

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

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

$n=\frac{2\sin \frac{\theta }{2}[\cos \frac{\theta }{2}+\sin \frac{\theta }{2}]}{2\cos \frac{\theta }{2}[\sin \frac{\theta }{2}+\cos \frac{\theta }{2}]}=\tan \frac{\theta }{2}$