Question

$\sqrt{\frac{sec\theta -1}{sec\theta +1}}+\sqrt{\frac{sec\theta +1}{sec\theta -1}}=2\cos ec\theta$

Answer 1

Prove the given expression.

$\sqrt{\frac{sec\theta -1}{sec\theta +1}}+\sqrt{\frac{sec\theta +1}{sec\theta -1}}=2\cos ec\theta$

Now, L.H.S. $\sqrt{\frac{sec\theta -1}{sec\theta +1}}+\sqrt{\frac{sec\theta +1}{sec\theta -1}}$

$$\begin{gathered} =\sqrt{\frac{{\displaystyle \frac{1}{\cos \theta }}-1}{{\displaystyle \frac{1}{\cos \theta }}+1}}+\sqrt{\frac{{\displaystyle \frac{1}{\cos \theta }}+1}{{\displaystyle \frac{1}{\cos \theta }}-1}} \\ =\sqrt{\frac{{\displaystyle \frac{1-\cos \theta }{\cos \theta }}}{{\displaystyle \frac{1+\cos \theta }{\cos \theta }}}}+\sqrt{\frac{{\displaystyle \frac{1+\cos \theta }{\cos \theta }}}{{\displaystyle \frac{1-\cos \theta }{\cos \theta }}}} \\ =\sqrt{\frac{1-\cos \theta }{\cos \theta }\times \frac{\cos \theta }{1+\cos \theta }}+\sqrt{\frac{1+\cos \theta }{\cos \theta }\times \frac{\cos \theta }{1-\cos \theta }} \\ =\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}+\sqrt{\frac{1+\cos \theta }{1-\cos \theta }} \\ =\frac{{\left(\sqrt{1-\cos \theta }\right)}^2+{\left(\sqrt{1+\cos \theta }\right)}^2}{\left(\sqrt{1+\cos \theta }\right)\left(\sqrt{1-\cos \theta }\right)} \\ =\frac{(1-\cos \theta )+(1+\cos \theta )}{\sqrt{1-{\cos }^2\theta }} \\ =\frac{1-\cos \theta +1+\cos \theta }{\sqrt{{\sin }^2\theta }} \\ =\frac{2}{\sqrt{{\sin }^2\theta }} \\ =\frac{2}{\sin \theta } \\ =2\cos ec\theta \end{gathered}$$

Hence, L.H.S.=R.H.S.