Question

$\sqrt{\frac{1+\sin \mathrm{\theta }}{1-\sin \mathrm{\theta }}}+\sqrt{\frac{1-\sin \mathrm{\theta }}{1+\sin \mathrm{\theta }}}=2\sec \mathrm{\theta }.$

Answer 1

Prove the given expression$\sqrt{\frac{1+\sin \mathrm{\theta }}{1-\sin \mathrm{\theta }}}+\sqrt{\frac{1-\sin \mathrm{\theta }}{1+\sin \mathrm{\theta }}}=2\sec \mathrm{\theta }$

$\sqrt{\frac{1+\sin \mathrm{\theta }}{1-\sin \mathrm{\theta }}}+\sqrt{\frac{1-\sin \mathrm{\theta }}{1+\sin \mathrm{\theta }}}$

$$\begin{gathered} Nowwemultipyinganddividingby1+\sin \theta \&1-\sin \theta \\ =\sqrt{\frac{1+\sin \mathrm{\theta }}{1-\sin \mathrm{\theta }}\times }\sqrt{\frac{1+\sin \mathrm{\theta }}{1+\sin \mathrm{\theta }}}+\sqrt{\frac{1-\sin \mathrm{\theta }}{1+\sin \mathrm{\theta }}\times }\sqrt{\frac{1-\sin \mathrm{\theta }}{1-\sin \mathrm{\theta }}} \end{gathered}$$

Now L.C.M case,

$=\frac{\sqrt{{\left(1+\sin \mathrm{\theta }\right)}^2}+\sqrt{{\left(1-\sin \mathrm{\theta }\right)}^2}}{\sqrt{\left(1-{\sin }^2\mathrm{\theta }\right)}}$

$=\frac{1+\sin \mathrm{\theta }+1-\sin \mathrm{\theta }}{\sqrt{1-{\sin }^2\mathrm{\theta }}}$

$=\frac{2}{\sqrt{{\cos }^2\mathrm{\theta }}}=\frac{2}{\cos \mathrm{\theta }}$

$=2\sec \mathrm{\theta }$ = R.H.S

Here, L.H.S = R.H.S Proved.