Question

Find the value of:
$\frac{1+\cos A-\sin A}{1+\cos A+\sin A}-\sec A+\tan A$

Answer 1

$\frac{1+\cos A-\sin A}{1+\cos A+\sin A}-\frac{1}{\cos A}+\frac{\sin A}{\cos A}\Rightarrow \frac{\left(\cos A\right)(1+\cos A-\sin A)-(1+\cos A+\sin A)+\sin A(1+\cos A+\sin A)}{\left(\cos A\right)(1+\cos A\sin A)}\Rightarrow \frac{{\sin }^2+{\cos }^2-1}{\left(\cos A\right)(1+\cos A\sin A)}=0\because [{\sin }^{2}x+{\cos }^{2}x=1]$