If $$\sin \theta = \frac{p^2 -1}{p^2 + 1}$$, then $$\cos \theta $$ is equal to:

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

We know that $$\sin ^2 \theta + \cos ^2 \theta = 1$$

=> $$ cos ^2 \theta = 1 - \sin^2 \theta $$

=> $$ cos \theta = \sqrt {1-sin^2\theta}$$

$$\cos \theta = \sqrt{1- (\frac{p^2 -1}{p^2 + 1})^2}$$

= $$\sqrt{1- (\frac{p^4 +1 -2p^2}{p^4+ 1 +2p^2})}$$

=$$\sqrt{ \frac{p^4+ 1 +2p^2 - p^4 -1 + 2p^2}{p^4+ 1 +2p^2})}$$

=$$\sqrt{ \frac{ 4p^2}{p^4+ 1 +2p^2}}$$

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

So , the answer would be Option a)$$\frac{2p}{1+p^2}$$.