Question

Prove that $\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}=\sec A\csc A+1.$

Answer 1

$L.H.S.$
$=\frac{{\sin }^{2}A}{\cos A(\sin A-\cos A)}-\frac{{\cos }^{2}A}{\sin A(\sin A-\cos A)}$
$=\frac{{\sin }^{3}A-{\cos }^{3}A}{\sin A\cos A(\sin A-\cos A)}$
$=\frac{{\sin }^{2}A+{\cos }^{2}A+\sin A\cos A}{\sin A\cos A}$
$=\left(\frac{1+\sin A\cos A}{\sin A\cos A}\right)$
$=\frac{1}{\sin A\cos A}+1=\sec A\csc A+1.=R.H.S$