Question

$\sqrt{\frac{1+\cos \mathrm{\theta }}{1-\cos \mathrm{\theta }}}+\sqrt{\frac{1-\cos \mathrm{\theta }}{1+\cos \mathrm{\theta }}}=2\mathrm{cosec}\mathrm{\theta }.$

Answer 1

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

Let L.H.S = R.H.S

Now L.H.S,

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

$$\begin{gathered} Nowmultiplying\&Dividingby1+\cos \theta and1-\cos \theta \\ =\sqrt{\frac{1+\cos \mathrm{\theta }}{1-\cos \mathrm{\theta }}}\times \sqrt{\frac{1+\cos \mathrm{\theta }}{1+\cos \mathrm{\theta }}}+\sqrt{\frac{1-\cos \mathrm{\theta }}{1+\cos \mathrm{\theta }}}\times \sqrt{\frac{1-\cos \mathrm{\theta }}{1-\cos \mathrm{\theta }}} \end{gathered}$$

Now L.C.M case,

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

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

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

$=\frac{2}{\mathrm{sin\theta }}=2\mathrm{cosec}\mathrm{\theta }$

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