If operatorname{cosec} theta = frac{p+q}{p-q} | vedclass

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

Question

(A) Given $\operatorname{cosec} 	heta = \frac{p+q}{p-q}$.We know that $\sin 	heta = \frac{p-q}{p+q}$.Using the formula $\sin 	heta = \frac{2 	an(\

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given $\operatorname{cosec} 	heta = \frac{p+q}{p-q}$.We know that $\sin 	heta = \frac{p-q}{p+q}$.Using the formula $\sin 	heta = \frac{2 	an(	heta/2)}{1 + 	an^2(	heta/2)}$,we have:$\frac{2 	an(	heta/2)}{1 + 	an^2(	heta/2)} = \frac{p-q}{p+q}$Applying componendo and dividendo:$\frac{1 + 	an^2(	heta/2) + 2 	an(	heta/2)}{1 + 	an^2(	heta/2) - 2 	an(	heta/2)} = \frac{(p+q) + (p-q)}{(p+q) - (p-q)}$$\frac{(1 + 	an(	heta/2))^2}{(1 - 	an(	heta/2))^2} = \frac{2p}{2q} = \frac{p}{q}$Taking the square root on both sides:$\frac{1 + 	an(	heta/2)}{1 - 	an(	heta/2)} = \sqrt{\frac{p}{q}}$Since $	an(\pi/4) = 1$,the expression becomes:$	an\left(\frac{\pi}{4} + \frac{	heta}{2}ight) = \sqrt{\frac{p}{q}}$Therefore,$\cot\left(\frac{\pi}{4} + \frac{	heta}{2}ight) = \frac{1}{	an(\pi/4 + 	heta/2)} = \sqrt{\frac{q}{p}}$.

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

Similar Questions

The expression $\frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A}$ can be written as:

Let $\alpha = \frac{1}{\sin 60^{\circ} \sin 61^{\circ}} + \frac{1}{\sin 62^{\circ} \sin 63^{\circ}} + \dots + \frac{1}{\sin 118^{\circ} \sin 119^{\circ}}$. Then the value of $\left(\frac{\operatorname{cosec} 1^{\circ}}{\alpha}\right)^2$ is $....$

If $\tan \beta = \frac{n \sin \alpha \cos \alpha}{1 - n \cos^2 \alpha}$,then $\tan (\alpha + \beta) \cdot \cot \alpha =$

If $\cos(x) + \cos^2(x) = 1$,then $\sin^2(x) + \sin^4(x)$ is equal to

$\cos ^2 76^{\circ}+\sin ^2 46^{\circ}+\sin 76^{\circ} \cos 46^{\circ} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module