Question

If $\sec \theta +\tan \theta =p$ then find the value of $\cos \theta$

Answer 1

$\sec \theta +\tan \theta =p$ .......$\left(1\right)$

$\Rightarrow (\sec \theta +\tan \theta )(\sec \theta -\tan \theta )=p(\sec \theta -\tan \theta )$

$\Rightarrow {\sec }^2\theta -{\tan }^2\theta =p(\sec \theta -\tan \theta )$

$\Rightarrow 1=p(\sec \theta -\tan \theta )$

$\Rightarrow \sec \theta -\tan \theta =\frac{1}{p}$ .......$\left(2\right)$

Adding $\left(1\right)$ and $\left(2\right)$ we get

$\Rightarrow \sec \theta +\tan \theta +\sec \theta -\tan \theta =p+\frac{1}{p}$

$\Rightarrow 2\sec \theta =\frac{p^2+1}{p}$

$\Rightarrow \sec \theta =\frac{p^2+1}{2p}$

$\Rightarrow \frac{1}{\cos \theta }=\frac{p^2+1}{2p}$

$\therefore \cos \theta =\frac{2p}{p^2+1}$