Question

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

Answer 1

$$\begin{gathered} \mathrm{Given}:\sec \theta +\tan \theta =p \\ \mathrm{Now},\frac{p^2-1}{p^2+1}=\frac{{\left(sec\theta +\tan \theta \right)}^2-1}{{\left(sec\theta +\tan \theta \right)}^2+1} \\ =\frac{se{c}^2\theta +{\tan }^2\theta +2sec\theta \tan \theta -1}{se{c}^2\theta +{\tan }^2\theta +2sec\theta \tan \theta +1} \\ =\frac{2{\tan }^2\theta +2\tan \theta sec\theta }{2se{c}^2\theta +2sec\theta \tan \theta } \\ =\frac{2\tan \theta \left(\tan \theta +sec\theta \right)}{2sec\theta (tan\theta +sec\theta )} \\ =\frac{\tan \theta }{sec\theta } \\ =\frac{\sin \theta }{\cos \theta }\times \frac{\cos \theta }{1}=\sin \theta \end{gathered}$$

Hence, proved.