If csc theta = frac{p + q}{p - q},then cot le | vedclass

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

Question

(B) Given,$\csc 	heta = \frac{p + q}{p - q}$.This implies $\frac{1}{\sin 	heta} = \frac{p + q}{p - q}$.Applying the componendo and dividendo rule:$\

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given,$\csc 	heta = \frac{p + q}{p - q}$.This implies $\frac{1}{\sin 	heta} = \frac{p + q}{p - q}$.Applying the componendo and dividendo rule:$\frac{1 + \sin 	heta}{1 - \sin 	heta} = \frac{(p + q) + (p - q)}{(p + q) - (p - q)} = \frac{2p}{2q} = \frac{p}{q}$.We know that $1 + \sin 	heta = (\cos \frac{	heta}{2} + \sin \frac{	heta}{2})^2$ and $1 - \sin 	heta = (\cos \frac{	heta}{2} - \sin \frac{	heta}{2})^2$.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 the 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(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,where $A = \frac{\pi}{4}$ and $B = \frac{	heta}{2}$:$	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 $\csc \theta = \frac{p + q}{p - q}$,then $\cot \left( \frac{\pi}{4} + \frac{\theta}{2} \right) = $

Similar Questions

An observer on the top of a cliff,$200\, m$ above the sea level,observes the angle of depression of two ships at anchor to be $45^{\circ}$ and $30^{\circ}$ respectively. Find the distance between the ships,if the line joining them stretches to the base of the cliff. (in $m$)

Evaluate: $\tan A + \cot (180^\circ + A) + \cot (90^\circ + A) + \cot (360^\circ - A)$

If $a \cos \theta + b \sin \theta = m$ and $a \sin \theta - b \cos \theta = n,$ then ${a^2} + {b^2} = $

If $\sin^2 \theta = \frac{x^2 + y^2 + 1}{2x}$,then $x$ must be

If $\theta$ is an acute angle and $\sin\theta = \frac{p - 6}{8 - p}$,then $p$ must satisfy:

Vedclass Test Series

Exam Paper Generator

Online Exam Module