Question

Solve:$\frac{1+\tan A}{1-\tan A}$= ?

Answer 1

Given,

$\frac{1+\tan A}{1-\tan A}=\frac{1+\frac{\sin A}{\cos A}}{1-\frac{\sin A}{\cos A}}=\frac{\frac{\cos A+\sin A}{\cos A}}{\frac{\cos A-\sin A}{\cos A}}=\frac{\cos A+\sin A}{\cos A-\sin A}=\frac{\cos A+\sin A}{\cos A-\sin A}\times \frac{\cos A+\sin A}{\cos A+\sin A}=\frac{{(\cos A+\sin A)}^2}{({\cos }^{2}A-{\sin }^{2}A)}=\frac{{\cos }^{2}A+{\sin }^{2}A+2\cos A\sin A}{{\cos }^{2}A-{\sin }^{2}A}=\frac{1+\sin 2A}{\cos 2A}[\because {\cos }^{2}A+{\sin }^{2}A=1,2\cos A\sin A=\sin 2A,{\cos }^{2}A-{\sin }^{2}A=\cos 2A]\therefore \frac{1+\tan A}{1-\tan A}=\frac{1+\sin 2A}{\cos 2A}.$