Question

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

Answer 1

$\frac{\cos 2A}{1-\sin 2A}$

$=\frac{{\cos }^{2}A-{\sin }^{2}A}{{\cos }^{2}A+{\sin }^{2}A-2\sin A\cos A}\{{\cos }^{2}A={\cos }^{2}A-{\sin }^{2}A,{\cos }^{2}A+{\sin }^{2}A=1and2\cos A\sin A=\sin 2A\}$

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

$=\frac{\left(\cos A\sin A\right)(\cos A-\sin A)}{(\cos A-\sin A{)}^2}$

$=\frac{\cos A+\sin A}{\cos A-\sin A}\{a^2-b^2=(a+b\left)\right(a-b\left)\right\}$

Dividing numerator and denominator by $\cos A$

$=\frac{1+\tan A}{1-\tan A}$