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

Answer 1

Toolbox:

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

LHS

$\large\frac{1+\sin \theta-\cos \theta}{1+\sin \theta+\cos \theta}$

$\large\frac{1-\cos \theta+\sin \theta}{1+\cos \theta+\sin \theta}=\frac{2\sin ^2\Large\frac{\theta}{2}+2\sin \Large\frac{\theta}{2}.\cos\large\frac{\theta}{2}}{2\cos ^2\Large\frac{\theta}{2}+2\sin \Large\frac{\theta}{2}.\cos\large\frac{\theta}{2}}$

$\Rightarrow \large\frac{2\sin \large\frac{\theta}{2}(\sin \Large\frac{\theta}{2}+\cos\Large\frac{\theta}{2})}{2\sin \large\frac{\theta}{2}(\sin \Large\frac{\theta}{2}+\cos\Large\frac{\theta}{2})}$

$\Rightarrow \tan\large\frac{\theta}{2}$

Hence proved