Question

Prove that : $\sqrt{\frac{1+\sin 2A}{1-\sin 2A}}=\tan (\frac{\pi }{4}+A).$

Answer 1

$Provethat:\sqrt{\frac{1+\sin 2A}{1-\sin 2A}}=\tan \left(\frac{\pi }{4}+A\right)$

L.H.S

$\sqrt{\frac{1+\sin 2A}{1-\sin 2A}}=\sqrt{\frac{{\left(\sin A+\cos A\right)}^2}{{\left(\cos A-\sin A\right)}^2}}=\frac{\sin A+\cos A}{-\sin A+\cos A}=\frac{1+\tan A}{1-\tan A}=\tan \left(\frac{\pi }{4}+A\right)$

Hence, proved.