Question

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

Answer 1

$\sec \theta +\tan \theta =p$

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

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

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

$\Rightarrow \frac{2\tan \theta (1+\tan \theta )}{2\sec \theta (1+\sec \theta )}\Rightarrow \frac{\tan \theta (\cos \theta +1)}{\sec \theta (\cos \theta +1)}$

$\Rightarrow \tan \theta$