If y = tan^{-1} left( frac{1 - cos 3x}{sin 3x | vedclass

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

Question

(C) Given,$y = 	an^{-1} \left( \frac{1 - \cos 3x}{\sin 3x} ight)$.Using trigonometric identities $1 - \cos 	heta = 2 \sin^2 \left( \frac{	heta}{2

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(C) Given,$y = 	an^{-1} \left( \frac{1 - \cos 3x}{\sin 3x} ight)$.Using trigonometric identities $1 - \cos 	heta = 2 \sin^2 \left( \frac{	heta}{2} ight)$ and $\sin 	heta = 2 \sin \left( \frac{	heta}{2} ight) \cos \left( \frac{	heta}{2} ight)$,we have:$y = 	an^{-1} \left( \frac{2 \sin^2 \left( \frac{3x}{2} ight)}{2 \sin \left( \frac{3x}{2} ight) \cos \left( \frac{3x}{2} ight)} ight)$$y = 	an^{-1} \left( \frac{\sin \left( \frac{3x}{2} ight)}{\cos \left( \frac{3x}{2} ight)} ight)$$y = 	an^{-1} \left( 	an \left( \frac{3x}{2} ight) ight)$$y = \frac{3x}{2}$Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{3x}{2} ight) = \frac{3}{2}$.

If $y = \tan^{-1} \left( \frac{1 - \cos 3x}{\sin 3x} \right)$,then $\frac{dy}{dx} = \ldots$

Similar Questions

The differential coefficient of ${\sec ^{ - 1}}\left( \frac{1}{{2{x^2} - 1}} \right)$ with respect to $\sqrt {1 - {x^2}} $ at $x = \frac{1}{2}$ is:

If $y = \tan^{-1}\left( \frac{x}{\sqrt{1 - x^2}} \right)$,then $\frac{dy}{dx} = $

If $y = \tan^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right)$,where $0 \leqslant x < \frac{\pi}{2}$,then find the value of $y'\left(\frac{\pi}{6}\right)$.

If $y=\tan ^{-1}\left[\frac{\sin ^3(2 x)-3 x^2 \sin (2 x)}{3 x \sin ^2(2 x)-x^3}\right]$,then $\frac{d y}{d x}=$

The derivative of $\sin ^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right)$ with respect to $\cos ^{-1} x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module