Question

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

Answer 1

$\frac{p^2-1}{p^2+1}$

$=\frac{{(\sec \theta +\tan \theta )}^2-1}{{(\sec \theta +\tan \theta )}^2+1}$

$\frac{{\sec }^2\theta +{\tan }^2\theta +2\sec \theta \tan \theta -1}{{\sec }^2\theta +{\tan }^2\theta +2\sec \theta \tan \theta +1}$

$=\frac{({\sec }^2\theta -1)+{\tan }^2\theta +2\sec \theta \tan \theta }{{\sec }^2\theta +({\tan }^2\theta +1)+2\sec \theta \tan \theta }$

$=\frac{{\tan }^2\theta +{\tan }^2\theta +2\sec \theta \tan \theta }{{\sec }^2\theta +{\sec }^2\theta +2\sec \theta \tan \theta }$

$=\frac{2{\tan }^2\theta +2\sec \theta \tan \theta }{2{\sec }^2\theta +2\sec \theta \tan \theta }$

$=\frac{2\tan \theta (\tan \theta +\sec \theta )}{2\sec \theta (\sec \theta +\tan \theta )}$

$=\frac{2\tan \theta }{2\sec \theta }$

$=\tan \theta \times \cos \theta$

$=\frac{\sin \theta }{\cos \theta }\times \cos \theta$

$=\sin \theta$