If $\operatorname{cosec} \theta + \cot \theta = p$,then prove that $\cos \theta = \frac{p^{2} - 1}{p^{2} + 1}$.

(N/A) Given,$\operatorname{cosec} \theta + \cot \theta = p$.
We know that $\operatorname{cosec} \theta = \frac{1}{\sin \theta}$ and $\cot \theta = \frac{\cos \theta}{\sin \theta}$.
Substituting these values,we get $\frac{1 + \cos \theta}{\sin \theta} = p$.
Squaring both sides,we get $\frac{(1 + \cos \theta)^{2}}{\sin^{2} \theta} = p^{2}$.
Using the identity $\sin^{2} \theta = 1 - \cos^{2} \theta = (1 - \cos \theta)(1 + \cos \theta)$,we have:
$\frac{(1 + \cos \theta)^{2}}{(1 - \cos \theta)(1 + \cos \theta)} = p^{2} \Rightarrow \frac{1 + \cos \theta}{1 - \cos \theta} = p^{2}$.
Applying the componendo and dividendo rule:
$\frac{p^{2} - 1}{p^{2} + 1} = \frac{(1 + \cos \theta) - (1 - \cos \theta)}{(1 + \cos \theta) + (1 - \cos \theta)}$.
Simplifying the expression:
$\frac{p^{2} - 1}{p^{2} + 1} = \frac{1 + \cos \theta - 1 + \cos \theta}{1 + \cos \theta + 1 - \cos \theta} = \frac{2 \cos \theta}{2} = \cos \theta$.
Thus,$\cos \theta = \frac{p^{2} - 1}{p^{2} + 1}$. Hence proved.