# Prove the:

Question:

Prove the: $\left|\sqrt{\frac{1-\sin x}{1+\sin x}}\right|+\left|\sqrt{\frac{1+\sin x}{1-\sin x}}\right|=-\frac{2}{\cos x}$, where $\frac{\pi}{2} Solution:$\mathrm{LHS}=\left|\sqrt{\frac{1-\sin x}{1+\sin x}}\right|+\left|\sqrt{\frac{1+\sin x}{1-\sin x}}\right|=\left|\sqrt{\frac{(1-\sin x)(1-\sin x)}{(1+\sin x)(1-\sin x)}}\right|+\left|\sqrt{\frac{(1+\sin x)(1+\sin x)}{(1-\sin x)(1+\sin x)}}\right|=\left|\sqrt{\frac{(1-\sin x)(1-\sin x)}{(1+\sin x)(1-\sin x)}}\right|+\left|\sqrt{\frac{(1+\sin x)(1+\sin x)}{(1-\sin x)(1+\sin x)}}\right|=\left|\sqrt{\frac{(1-\sin x)^{2}}{1-\sin ^{2} x}}\right|+\left|\sqrt{\frac{(1+\sin x)^{2}}{1-\sin ^{2} x}}\right|=\left|\sqrt{\frac{(1-\sin x)^{2}}{\cos ^{2} x}}\right|+\left|\sqrt{\frac{(1+\sin x)^{2}}{\cos ^{2} x}}\right|=\left|\frac{1-\sin x}{\cos x}\right|+\left|\frac{1+\sin x}{\cos x}\right|=\left|\frac{1-\sin x+1+\sin x}{\cos x}\right|=\left|\frac{2}{\cos x}\right|=-\frac{2}{\cos x} \quad\left[\because \frac{\pi}{2}

= RHS

Hence proved.