Solve the following

Given, $\quad \operatorname{cosec} \theta+\cot \theta=p$

$\Rightarrow \quad \frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}=p$ $\left[\because \operatorname{cosec} \theta=\frac{1}{\sin \theta}\right.$ and $\left.\cot \theta=\frac{\cos \theta}{\sin \theta}\right]$

$\Rightarrow$ $\frac{1+\cos \theta}{\sin \theta}=\frac{p}{1}$ 

$\Rightarrow$ $\frac{(1+\cos \theta)^{2}}{\sin ^{2} \theta}=\frac{p^{2}}{1}$ [take square on both sides]

$\Rightarrow \quad \frac{1+\cos ^{2} \theta+2 \cos \theta}{\sin ^{2} \theta}=\frac{p^{2}}{1}$

Using componendo and dividendo rule, we get

$\frac{\left(1+\cos ^{2} \theta+2 \cos \theta\right)-\sin ^{2} \theta}{\left(1+\cos ^{2} \theta+2 \cos \theta\right)+\sin ^{2} \theta}=\frac{p^{2}-1}{p^{2}+1}$

$\Rightarrow$ $\frac{1+\cos ^{2} \theta+2 \cos \theta-\left(1-\cos ^{2} \theta\right)}{1+2 \cos \theta+\left(\cos ^{2} \theta+\sin ^{2} \theta\right)}=\frac{p^{2}-1}{p^{2}+1}$  $\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$

$\Rightarrow$ $\frac{2 \cos ^{2} \theta+2 \cos \theta}{2+2 \cos \theta}=\frac{p^{2}-1}{p^{2}+1}$

$\Rightarrow$ $\frac{2 \cos \theta(\cos \theta+1)}{2(\cos \theta+1)}=\frac{p^{2}-1}{p^{2}+1}$

$\therefore$ $\cos \theta=\frac{p^{2}-1}{p^{2}+1}$ Hence proved.