If $cos \theta = \frac{p^2-1}{p^2+1}, 0° < θ < 90°,$ then cosecθ is equal to :

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

cos θ = \(\frac{p² - 1}{p² + 1}\)

{ cos A = \(\frac{B}{H}\) }

B = p² - 1  &  H = p² + 1

By using pythagoras theorem,

P² + B² =  H²

P² + ( p² - 1 )² = ( p² + 1 )²

P = \(\sqrt { ( p² + 1 )² -( p² - 1 )²   }\)

P = \(\sqrt { 4p²   }\)

P = 2p

Now,

cosec θ = \(\frac{H}{P}\)

= \(\frac{ p² + 1}{2p}\)