Prove that

Solution:

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

Proof: Consider, L.H.S $=\sqrt{\frac{1+\sin x}{1-\sin x}}$

Multiply and divide L.H.S by $\sqrt{1+\sin x}$

$=\sqrt{\frac{1+\sin x}{1-\sin x}} \times \frac{\sqrt{1+\sin x}}{\sqrt{1+\sin x}}=\frac{1+\sin x}{\sqrt{1-\sin ^{2} x}}$

$=\frac{1+\sin x}{\cos x}=\frac{1+2 \cos \frac{x}{2} \sin \frac{x}{2}}{\cos x}\left(\because 2 \cos \frac{x}{2} \sin \frac{x}{2}=\sin x\right)$

$=\frac{\cos ^{2} \frac{x}{2}+\sin ^{2} \frac{x}{2}+2 \cos \frac{x}{2} \sin \frac{x}{2}}{\cos x}\left(\because \cos ^{2} x+\sin ^{2} x=1\right)$

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

$=\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}} \times \frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\left(\because x^{2}+y^{2}=(x+y)(x-y)\right)$

$=\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}$

Multiply and divide the above with $\cos \frac{x}{2}$

$=\frac{1+\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}}{1-\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}}$

Here, since $\tan \frac{\pi}{4}=1$

Here, since $\tan \frac{\pi}{4}=1$

$\sqrt{\frac{1+\sin x}{1-\sin x}}=\frac{\tan \frac{\pi}{4}+\tan \frac{x}{2}}{1-\tan \frac{\pi}{4} \tan \frac{\pi}{2}}=\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)=$ R.H.S

Since, L.H.S = R.H.S, Hence proved.