Question

Prove the: $\left|\sqrt{{\displaystyle \frac{1-\sin x}{1+\sin x}}}\right|+\left|\sqrt{{\displaystyle \frac{1+\sin x}{1-\sin x}}}\right|=-\frac{2}{\cos x},\mathrm{where}\frac{\pi }{2}<x<\pi$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\left|\sqrt{{\displaystyle \frac{1-\sin x}{1+\sin x}}}\right|+\left|\sqrt{{\displaystyle \frac{1+\sin x}{1-\sin x}}}\right| \\ =\left|\sqrt{\frac{\left(1-\sin x\right)\left(1-\sin x\right)}{\left(1+\sin x\right)\left(1-\sin x\right)}}\right|+\left|\sqrt{\frac{\left(1+\sin x\right)\left(1+\sin x\right)}{\left(1-\sin x\right)\left(1+\sin x\right)}}\right| \\ =\left|\sqrt{\frac{\left(1-\sin x\right)\left(1-\sin x\right)}{\left(1+\sin x\right)\left(1-\sin x\right)}}\right|+\left|\sqrt{\frac{\left(1+\sin x\right)\left(1+\sin x\right)}{\left(1-\sin x\right)\left(1+\sin x\right)}}\right| \\ =\left|\sqrt{\frac{{\left(1-\sin x\right)}^2}{1-{\sin }^{2}x}}\right|+\left|\sqrt{\frac{{\left(1+\sin x\right)}^2}{1-{\sin }^{2}x}}\right| \\ =\left|\sqrt{\frac{{\left(1-\sin x\right)}^2}{{\cos }^{2}x}}\right|+\left|\sqrt{\frac{{\left(1+\sin x\right)}^2}{{\cos }^{2}x}}\right| \\ =\left|\frac{1-\sin x}{\cos x}\right|+\left|\frac{1+\sin x}{\cos x}\right| \\ =\left|\frac{1-\sin x+1+\sin x}{\cos x}\right| \\ =\left|\frac{2}{\cos x}\right| \\ =-\frac{2}{\cos x}\left[\because \frac{\mathrm{\pi }}{2}<x<\mathrm{\pi }\mathrm{and}\mathrm{in}\mathrm{the}\mathrm{second}\mathrm{quadrant},\cos x\mathrm{is}\mathrm{negative}\right] \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$