Correct option is: A) \(\frac{p^2-1}{p^2+1}\)

We have 

sec \(\theta\) + tan \(\theta\) = P ....(1)

On multiplying both sides of equation (1) by(sec \(\theta\) - tan \(\theta\) ), we get

(sec \(\theta\) + tan \(\theta\) ) (sec \(\theta\) - tan \(\theta\) ) = P ( sec \(\theta\) - tan \(\theta\) )

= P ( sec \(\theta\) - tan \(\theta\)) = \(\sec^2\, \theta - tan^2 \, \theta = 1\)  (\(\because\) \(\sec^2\, \theta - tan^2 \, \theta = 1\))

= sec \(\theta\) - tan \(\theta\) = \(\frac 1P\) ....(2)

By adding equations (1) & (2), we get

(sec \(\theta\) + tan \(\theta\)) + ( sec \(\theta\) - tan \(\theta\)) = P + \(\frac 1P\)

= 2 sec \(\theta\) = \(\frac {P^2+1}{P}\) 

= sec \(\theta\) = \(\frac {P^2+1}{2P}\) ....(3)

By subtracting equation (2) from equation (1), we get

(sec \(\theta\) + tan \(\theta\)) - ( sec \(\theta\) - tan \(\theta\)) = P - \(\frac 1P\)

= 2 tan \(\theta\) = \(\frac {P^2-1}{P}\) 

= tan \(\theta\) = \(\frac {P^2-1}{2P}\) ....(4)

Now, sin \(\theta\) = \(\frac {\frac {sin\, \theta}{cos\, \theta}}{\frac 1{cos\, \theta}} = \frac {tan\, \theta}{sec\, \theta}= \frac {\frac {P^2-1}{2P}}{\frac {P^2+1}{2P}} = \frac {P^2-1}{P^2+1}\)  (From (3) & (4)

Correct option is: A) \(\frac{p^2-1}{p^2+1}\)