Question

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

Answer 1

$sin\theta =\frac{2tan\frac{\theta }{2}}{1+ta{n}^2\frac{\theta }{2}}$

$cos\theta =\frac{1-ta{n}^2\frac{\theta }{2}}{1+ta{n}^2\frac{\theta }{2}}$

Substituting this in the above equation we get,$\frac{1+ta{n}^2\frac{\theta }{2}+2tan\frac{\theta }{2}-1+ta{n}^2\frac{\theta }{2}}{1+ta{n}^2\frac{\theta }{2}+2tan\frac{\theta }{2}+1-ta{n}^2\frac{\theta }{2}}$

$⟹\frac{2tan\frac{\theta }{2}+2ta{n}^2\frac{\theta }{2}}{2+2tan\frac{\theta }{2}}=tan\frac{\theta }{2}$