sqrt{frac{1 - sin A}{1 + sin A}} = | vedclass

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

Question

(D) We have $\sqrt{\frac{1 - \sin A}{1 + \sin A}}$.Using the identity $\sin A = \cos \left( \frac{\pi}{2} - A \right)$,we get $\sqrt{\frac{1 - \cos \l

Vedclass · Accepted Answer

$	an \left( \frac{\pi}{4} - \frac{A}{2} ight)$

Vedclass · Answer

(D) We have $\sqrt{\frac{1 - \sin A}{1 + \sin A}}$.Using the identity $\sin A = \cos \left( \frac{\pi}{2} - A ight)$,we get $\sqrt{\frac{1 - \cos \left( \frac{\pi}{2} - A ight)}{1 + \cos \left( \frac{\pi}{2} - A ight)}}$.Applying the half-angle formulas $1 - \cos 	heta = 2 \sin^2 \left( \frac{	heta}{2} ight)$ and $1 + \cos 	heta = 2 \cos^2 \left( \frac{	heta}{2} ight)$,we have:$\sqrt{\frac{2 \sin^2 \left( \frac{\pi}{4} - \frac{A}{2} ight)}{2 \cos^2 \left( \frac{\pi}{4} - \frac{A}{2} ight)}} = \sqrt{	an^2 \left( \frac{\pi}{4} - \frac{A}{2} ight)} = 	an \left( \frac{\pi}{4} - \frac{A}{2} ight)$.

$\sqrt{\frac{1 - \sin A}{1 + \sin A}} = $

Similar Questions

If $\left(\frac{\sin 3 \theta}{\sin \theta}\right)^2-\left(\frac{\cos 3 \theta}{\cos \theta}\right)^2=a \cos b \theta$,then $a: b=$

If $\cot \frac{2x}{3} + \tan \frac{x}{3} = \operatorname{cosec} \frac{kx}{3}$, then the value of $k$ is

$f(x) = 3 \sin x - 4 \sin^3 x$,find the value of $f\left(\frac{\pi}{3}\right)$.

$\sin 4\theta$ can be written as

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 $1$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module