If csc theta = frac{p + q}{p - q} (p neq q ne | vedclass

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

Question

(B) Given $\csc 	heta = \frac{p + q}{p - q}$,so $\sin 	heta = \frac{p - q}{p + q}$.Using the identity $\cot \left( \frac{\pi}{4} + \frac{	heta}{2}

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given $\csc 	heta = \frac{p + q}{p - q}$,so $\sin 	heta = \frac{p - q}{p + q}$.Using the identity $\cot \left( \frac{\pi}{4} + \frac{	heta}{2} ight) = \frac{\cot \frac{\pi}{4} \cot \frac{	heta}{2} - 1}{\cot \frac{\pi}{4} + \cot \frac{	heta}{2}} = \frac{\cot \frac{	heta}{2} - 1}{\cot \frac{	heta}{2} + 1}$.This simplifies to $\frac{\cos \frac{	heta}{2} - \sin \frac{	heta}{2}}{\cos \frac{	heta}{2} + \sin \frac{	heta}{2}}$.Squaring this expression,we get $\frac{\cos^2 \frac{	heta}{2} + \sin^2 \frac{	heta}{2} - 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}}{\cos^2 \frac{	heta}{2} + \sin^2 \frac{	heta}{2} + 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}} = \frac{1 - \sin 	heta}{1 + \sin 	heta}$.Substituting $\sin 	heta = \frac{p - q}{p + q}$:$\frac{1 - \frac{p - q}{p + q}}{1 + \frac{p - q}{p + q}} = \frac{p + q - p + q}{p + q + p - q} = \frac{2q}{2p} = \frac{q}{p}$.Taking the square root,we get $\left| \cot \left( \frac{\pi}{4} + \frac{	heta}{2} ight) ight| = \sqrt{\frac{q}{p}}$.

If $\csc \theta = \frac{p + q}{p - q}$ $(p \neq q \neq 0)$,then $\left| \cot \left( \frac{\pi}{4} + \frac{\theta}{2} \right) \right|$ is equal to

Similar Questions

If $\cos ^3 x \sin 4 x = \sum_{r=0}^{n} a_{r} \sin rx$ for all $x \in R$,then $a_3+a_5 : a_1+a_7 = $

$\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 8 \theta}}} = $

$\cos^3 \frac{\pi}{8} \cos \frac{3\pi}{8} + \sin^3 \frac{\pi}{8} \sin \frac{3\pi}{8} = $

If $\cos 3\theta = \alpha \cos \theta + \beta \cos^3 \theta$,then $(\alpha, \beta) = $

$\tan 20^\circ \cot 10^\circ \cot 50^\circ$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module