Question

If sec $\theta +\tan \theta =p,$ prove that $\sin \theta =\frac{p^2-1}{p^2+1}$

Answer 1

Given $\text{sec}\theta +\tan \theta =p$

$⟹\text{sec}\theta -\tan \theta =\frac{1}{p}$

So $2\text{sec}\theta =p+\frac{1}{p}⟹\text{sec}\theta =\frac{1}{2}(p+\frac{1}{p})=\frac{p^2+1}{2p}$

and also $2\tan \theta =p-\frac{1}{p}⟹\tan \theta =\frac{1}{2}(p-\frac{1}{p})=\frac{p^2-1}{2p}$

$\sin \theta =\frac{\tan \theta }{\text{sec}\theta }=\frac{\frac{p^2-1}{2p}}{\frac{p^2+1}{2p}}=\frac{p^2-1}{p^2+1}$