Question

Prove the following identities (1-17)

$\left|\sqrt{\frac{1-\sin \mathrm{\theta }}{1+\sin \mathrm{\theta }}}\right|+\left|\sqrt{\frac{1+\sin \mathrm{\theta }}{1-\sin \mathrm{\theta }}}\right|=-\frac{2}{\cos \mathrm{\theta }},\mathrm{where}\frac{\pi }{2}<\mathrm{\theta }<\pi$

Answer 1

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