Question

Prove that

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

Answer 1

$\frac{\sin A\tan A}{1-\cos A}=\frac{{\sin }^{2}A}{\cos A(1-\cos A)}$

$=\frac{1-{\cos }^{2}A}{\cos A(1-\cos A)}$

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

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