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

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

Question

(D) Given that,$y = 	an^{-1}\left(\frac{\sin x + \cos x}{\cos x - \sin x}ight)$.Dividing the numerator and denominator by $\cos x$,we get:$y = 	an

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) Given that,$y = 	an^{-1}\left(\frac{\sin x + \cos x}{\cos x - \sin x}ight)$.Dividing the numerator and denominator by $\cos x$,we get:$y = 	an^{-1}\left(\frac{	an x + 1}{1 - 	an x}ight)$.Using the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,where $A = \pi/4$ and $B = x$:$y = 	an^{-1}\left(	an\left(\frac{\pi}{4} + xight)ight) = \frac{\pi}{4} + x$.Now,differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{4} + xight) = 0 + 1 = 1$.

If $y = \tan^{-1}\left(\frac{\sin x + \cos x}{\cos x - \sin x}\right)$,then $\frac{dy}{dx}$ is equal to

Similar Questions

If $y=\cos ^{-1}\left(\frac{6 x-2 x^2-4}{2 x^2-6 x+5}\right)$, then $\frac{d y}{d x}=$

Derivative of $\tan ^{-1} \sqrt{\frac{1-x}{1+x}}$ with respect to $\cos ^{-1}\left(4 x^3-3 x\right)$ is

The differential coefficient of ${\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + {x^2}} - 1}}{x}} \right)$ with respect to ${\tan ^{ - 1}}x$ is

If $y = \cos^{-1} \left( \frac{1-4^x}{1+4^x} \right)$, then find $\frac{dy}{dx}$ at $x = 1$.

If $y=\cos ^{-1}\left(\frac{a^2}{\sqrt{x^4+a^4}}\right)$,then $\frac{d y}{d x}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module