frac{d}{dx} sqrt{frac{1 - sin 2x}{1 + sin 2x} | vedclass

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

Question

(B) Let $y = \sqrt{\frac{1 - \sin 2x}{1 + \sin 2x}}$.Using the identities $1 = \sin^2 x + \cos^2 x$ and $\sin 2x = 2 \sin x \cos x$,we have:$y = \sqrt

Vedclass · Accepted Answer

$-\sec^2 \left( \frac{\pi}{4} - x \right)$

Vedclass · Answer

(B) Let $y = \sqrt{\frac{1 - \sin 2x}{1 + \sin 2x}}$.Using the identities $1 = \sin^2 x + \cos^2 x$ and $\sin 2x = 2 \sin x \cos x$,we have:$y = \sqrt{\frac{\cos^2 x + \sin^2 x - 2 \sin x \cos x}{\cos^2 x + \sin^2 x + 2 \sin x \cos x}} = \sqrt{\frac{(\cos x - \sin x)^2}{(\cos x + \sin x)^2}} = \frac{\cos x - \sin x}{\cos x + \sin x}$.Dividing numerator and denominator by $\cos x$:$y = \frac{1 - 	an x}{1 + 	an x} = 	an \left( \frac{\pi}{4} - x ight)$.Now,differentiating with respect to $x$:$\frac{dy}{dx} = \sec^2 \left( \frac{\pi}{4} - x ight) \cdot \frac{d}{dx} \left( \frac{\pi}{4} - x ight) = \sec^2 \left( \frac{\pi}{4} - x ight) \cdot (-1) = -\sec^2 \left( \frac{\pi}{4} - x ight)$.

$\frac{d}{dx} \sqrt{\frac{1 - \sin 2x}{1 + \sin 2x}} = $

Similar Questions

If the function $g(x)$ is defined by $g(x) = \frac{x^{200}}{200} + \frac{x^{199}}{199} + \frac{x^{198}}{198} + \dots + \frac{x^2}{2} + x + 5$,then find $g'(0)$.

If $y = \log \tan \sqrt{x}$,then the value of $\frac{dy}{dx}$ is

Let $f: R \rightarrow R$ be defined by $f(x) = \log \left[e^x \left(\frac{x-2}{x+2}\right)^{3/4}\right]$. Find the value of $f'(0)$.

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

The differential coefficient of $\frac{{\tan^{-1}x}}{{1 + \tan^{-1}x}}$ with respect to $\tan^{-1}x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module