Question

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

Answer 1

$\frac{1+\sin A-\cos A}{1+\sin A+\cos A}$

We can write

$\Rightarrow \cos A=2{\cos }^2\frac{A}{2}-1$ $\Rightarrow \cos A+1=2{\cos }^2\frac{A}{2}$

$\Rightarrow \cos A=1-2{\sin }^2\frac{A}{2}$ $\Rightarrow 1-\cos A=2{\sin }^2\frac{A}{2}$

$\Rightarrow \sin A=2\sin \frac{A}{2}\cos \frac{A}{2}$

Substituting these values in the given expression we get,

$\Rightarrow \frac{1+\sin A-\cos A}{1+\sin A+\cos A}=\frac{(1-\cos A)+\sin A}{(1+\cos A)+\sin A}$

$\Rightarrow \frac{1+\sin A-\cos A}{1+\sin A+\cos A}=\frac{2{\sin }^2\frac{A}{2}+2\sin \frac{A}{2}\cos \frac{A}{2}}{2{\cos }^2\frac{A}{2}+2\sin \frac{A}{2}\cos \frac{A}{2}}$

$\Rightarrow \frac{1+\sin A-\cos A}{1+\sin A+\cos A}=\frac{2\sin \frac{A}{2}(\sin \frac{A}{2}+\cos \frac{A}{2})}{2\cos \frac{A}{2}(\sin \frac{A}{2}+\sin \frac{A}{2})}$

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