Question

Root sec teetha -1 / sec teetha +1 + root sec teetha+1 / sec teetha -1 = 2 cosec teetha

Answer 1

Dear student
$$\begin{gathered} \sqrt{\frac{sec\theta -1}{sec\theta +1}}+\sqrt{\frac{sec\theta +1}{sec\theta -1}} \\ =\sqrt{\frac{sec\theta -1}{sec\theta +1}\times \frac{sec\theta -1}{sec\theta -1}}+\sqrt{\frac{sec\theta +1}{sec\theta -1}\times \frac{sec\theta +1}{sec\theta +1}} \\ =\frac{sec\theta +1}{\sqrt{se{c}^2\theta -1}}+\frac{sec\theta -1}{\sqrt{se{c}^2\theta -1}} \\ =\frac{sec\theta +1}{\tan \theta }+\frac{sec\theta -1}{\tan \theta }\left[Asse{c}^{2}x=1+{\tan }^{2}x\right] \\ =\frac{sec\theta +1+sec\theta -1}{\tan \theta } \\ =\frac{2sec\theta }{\tan \theta } \\ =\frac{2}{\cos \theta }\times \frac{\cos \theta }{\sin \theta } \\ =2\cos ec\theta \end{gathered}$$
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