Question

Prove that $\frac{1+\sin A-\cos A}{cosA+\sin A-1}=\tan \frac{A}{2}$.

Answer 1

consider the following equation.

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

Solve LHS

$\frac{1+\sin A-\cos A}{1-\sin A+\cos A}$=$\frac{\csc \theta +\cot \theta -1}{1-\csc \theta +\cot \theta }$

$=\frac{\sec A+tanA-({\csc }^{2}A-{\cot }^{2}A)}{(1-\csc A+\cot A)}$

$=\frac{(\csc A+\cot A)-(\csc A-\cot A)(\csc A+\cot A)}{\csc A+\cot A+1}$

$=\frac{(\csc A+\cot A)(1-(\csc A-\cot A))}{(1-\csc A+\cot A)}$

$=\frac{(\csc A+\cot A)(1-(\csc A-\cot A))}{(1-\csc A+\cot A)}$

$=(\csc A+\cot A)$

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

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

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

$=\tan \frac{A}{2}$

LHS=RHS

Hence, proved