Correct Answer: $\cos \theta$

Solution : Given, $\operatorname{cosec} \theta+\cot \theta=p$
Squaring both sides, we get,
$\operatorname{cosec}^2 \theta+\cot^2 \theta+2\operatorname{cosec} \theta\cot \theta=p^2$ ----------------------(1)
Adding 1 on both sides of equation (1),
$\operatorname{cosec}^2 \theta+(\cot^2 \theta+1)+2\operatorname{cosec} \theta\cot \theta=p^2+1$
$⇒2\operatorname{cosec}^2 \theta+2\operatorname{cosec} \theta\cot \theta=p^2+1$ ---------------------------(2)
Subtracting 1 from both sides of equation (1),
$(\operatorname{cosec}^2-1) \theta+\cot^2 \theta+2\operatorname{cosec} \theta\cot \theta=p^2-1$
$⇒2\cot^2 \theta+2\operatorname{cosec} \theta\cot \theta=p^2-1$-------------------------------- (3)
Dividing equation (3) by equation (2),
$\frac{2\cot^2 \theta+2\operatorname{cosec} \theta\cot \theta}{2\operatorname{cosec}^2 \theta+2\operatorname{cosec} \theta\cot \theta}=\frac{p^2-1}{p^2+1}$
$⇒\frac{2\cot \theta(\cot \theta +\operatorname{cosec} \theta)}{2\operatorname{cosec} \theta(\cot \theta + \operatorname{cosec} \theta)}=\frac{p^2-1}{p^2+1}$
$⇒\frac{\cot \theta}{\operatorname{cosec} \theta}=\frac{p^2-1}{p^2+1}$
$\therefore\frac{p^2-1}{p^2+1}=\cos \theta$
Hence, the correct answer is $\cos \theta$.