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

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

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

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 = \(\frac{H}{P}\) = \(\frac{ p² + 1}{2p}\)