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}}$.

List-$I$	List-$II$
$A$. The period of $\sin^2 x$ is	$I$. $\frac{2\pi}{3}$
$B$. Maximum value of $\frac{\pi}{3}(\sqrt{3}\cos 3x + \sin 3x)$	$II$. $12\pi$
$C$. The period of $\sin \frac{x}{3} + \cos \frac{x}{2}$ is	$III$. $\frac{\pi}{2}$
$D$. Intersection points of $y=\|\sin x\|$ and $y=1$ in $(0, \pi)$	$IV$. $\frac{3\pi}{2}$
	$V$. $\pi$

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

Similar Questions

Let $S$ be the set of all $\alpha \in \mathbb{R}$ such that the equation $\cos 2x + \alpha \sin x = 2\alpha - 7$ has a solution. Then $S$ is equal to

The equation $2 \cos ^{2} \left( \frac{x}{2} \right) \sin ^{2} x = x^{2} + \frac{1}{x^{2}}$ for $0 \leq x \leq \frac{\pi}{2}$ has

If $A$ and $B$ are values such that $(A+B)$ and $(A-B)$ are not odd multiples of $\frac{\pi}{2}$ and $2 \tan (A+B)=3 \tan (A-B)$,then $\sin 2A$ is equal to:

The sum of the solutions of the equation $\cos x \cos \left(\frac{\pi}{3}-x\right) \cos \left(\frac{\pi}{3}+x\right)=\frac{1}{4}$ in the interval $(0, 2\pi)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module