Question

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

Answer 1

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

squaring on both sides and using the trigonometric identities, we get,

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

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

using $1={\cos }^{2}A+{\sin }^{2}A$ and $\tan A=\frac{\sin A}{\cos A}$

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

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

Hence proved.