If cosec θ = + q - q ( ≠ q, p≠ 0),then | cot( π 4+ θ 2)| is equals to:

$$\text{cosec } \theta = \frac{ \text{p} + \text{q} }{ \text{p} - \text{q} } \Rightarrow \frac{ 1 }{ \text{sin } \theta } = \frac{ \text{p} + \text{q} }{ \text{p} - \text{q} }$$cosec θ=p−qp+q​⇒sin θ1​=p−qp+q​

let $\text{tan} ( \frac{ \theta }{ 2 } ) = \text{t } \text{then } \frac{ 1 + {\text{t}}^{2} }{ \text{2t} } = \frac{ \text{p} + \text{q} }{ \text{p} - \text{q} }$tan(2θ​)=t then 2t1+t2​=p−qp+q​ ${\because \mathrm{sin}2\theta =\frac{2\mathrm{tan}\theta }{1+{\mathrm{tan}}^{2}\theta }}$∵sin2θ=1+tan2θ2tanθ​

Using componendo and dividendo rule, we get

$$\Rightarrow \frac{ 1 + {\text{t}}^{2} + \text{2t} }{ 1 + {\text{t}}^{2} - \text{2t} } = \frac{ \text{p} + \text{q} + \text{p} - \text{q} }{ \text{p} + \text{q} - \text{p} + \text{q} }$$⇒1+t2−2t1+t2+2t​=p+q−p+qp+q+p−q​

$$\Rightarrow { ( \frac{ 1 + \text{t} }{ 1 - \text{t} } ) }^{2} = \frac{ \text{p} }{ \text{q} }$$⇒(1−t1+t​)2=qp​

$\Rightarrow |\frac{1-t}{1+t}| =\sqrt{\frac{q}{p}}=|cot(\frac{\pi }{4}+\theta )|$⇒∣1+t1−t​∣=pq​​=∣cot(4π​+θ)∣ ${\because \mathrm{tan}(\frac{\pi }{4}+\theta )=\frac{1+\mathrm{tan}\theta }{1-\mathrm{tan}\theta }}$∵tan(4π​+θ)=1−tanθ1+tanθ​