If text{cosec} theta = frac{p + q}{p - q},the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given,$	ext{cosec} 	heta = \frac{p + q}{p - q}$.$\frac{1}{\sin 	heta} = \frac{p + q}{p - q}$.Applying componendo and dividendo:$\frac{1 + \sin

Vedclass · Accepted Answer

$\sqrt{\frac{q}{p}}$

Vedclass · Answer

(B) Given,$	ext{cosec} 	heta = \frac{p + q}{p - q}$.$\frac{1}{\sin 	heta} = \frac{p + q}{p - q}$.Applying componendo and dividendo:$\frac{1 + \sin 	heta}{1 - \sin 	heta} = \frac{(p + q) + (p - q)}{(p + q) - (p - q)} = \frac{2p}{2q} = \frac{p}{q}$.We know that $1 + \sin 	heta = \left( \cos \frac{	heta}{2} + \sin \frac{	heta}{2} ight)^2$ and $1 - \sin 	heta = \left( \cos \frac{	heta}{2} - \sin \frac{	heta}{2} ight)^2$.So,$\left( \frac{\cos \frac{	heta}{2} + \sin \frac{	heta}{2}}{\cos \frac{	heta}{2} - \sin \frac{	heta}{2}} ight)^2 = \frac{p}{q}$.Dividing numerator and denominator by $\cos \frac{	heta}{2}$,we get:$\left( \frac{1 + 	an \frac{	heta}{2}}{1 - 	an \frac{	heta}{2}} ight)^2 = \frac{p}{q}$.Using the formula $	an \left( \frac{\pi}{4} + \frac{	heta}{2} ight) = \frac{1 + 	an \frac{	heta}{2}}{1 - 	an \frac{	heta}{2}}$,we have:$	an^2 \left( \frac{\pi}{4} + \frac{	heta}{2} ight) = \frac{p}{q}$.Taking the reciprocal:$\cot^2 \left( \frac{\pi}{4} + \frac{	heta}{2} ight) = \frac{q}{p}$.Therefore,$\cot \left( \frac{\pi}{4} + \frac{	heta}{2} ight) = \sqrt{\frac{q}{p}}$.

If $\text{cosec} \theta = \frac{p + q}{p - q}$,then $\cot \left( \frac{\pi}{4} + \frac{\theta}{2} \right) = $

Similar Questions

If $\tan A < 0$ and $\tan 2A = -\frac{4}{3}$,then $\cos 6A =$

$\tan A + 2 \tan 2A + 4 \tan 4A + 8 \cot 8A = $

Prove that $\frac{\sin x - \sin 3x}{\sin^2 x - \cos^2 x} = 2 \sin x$.

If $\tan \theta + \tan \left(\theta + \frac{\pi}{3}\right) + \tan \left(\theta + \frac{2\pi}{3}\right) = 3$,then which of the following is equal to $1$?

If $\tan A = \frac{1}{2},$ then $\tan 3A = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module