If cosec  θ = p + q   p - q   p ≠ q, p ≠ 0 ,   then cot…

$\text{cosec }\theta =\frac{\text{p}+\text{q}}{\text{p}-\text{q}}\Rightarrow \frac{1}{\text{sin }\theta }=\frac{\text{p}+\text{q}}{\text{p}-\text{q}}$

let $\text{tan}\left(\frac{\theta }{2}\right)=\text{t }\text{then }\frac{1+{\text{t}}^2}{\text{2t}}=\frac{\text{p}+\text{q}}{\text{p}-\text{q}}$ $\left\{\because \sin 2\theta =\frac{2\tan \theta }{1+{\tan }^2\theta }\right\}$

Using componendo and dividendo rule, we get

$\Rightarrow \frac{1+{\text{t}}^2+\text{2t}}{1+{\text{t}}^2-\text{2t}}=\frac{\text{p}+\text{q}+\text{p}-\text{q}}{\text{p}+\text{q}-\text{p}+\text{q}}$  

$\Rightarrow {\left(\frac{1+\text{t}}{1-\text{t}}\right)}^2=\frac{\text{p}}{\text{q}}$

$\Rightarrow \left|\frac{1-t}{1+t}\right|=\sqrt{\frac{q}{p}}=\left|\mathrm{c}\mathrm{o}\mathrm{t}\right(\frac{\pi }{4}+\theta \left)\right|$ $\left\{\because \tan \left(\frac{\mathrm{\pi }}{4}+\theta \right)=\frac{1+\tan \theta }{1-\tan \theta }\right\}$