If y = tan^{-1} sqrt{frac{1 + sin 2x}{1 - sin | vedclass

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

Question

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

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Given $y = 	an^{-1} \sqrt{\frac{1 + \sin 2x}{1 - \sin 2x}}$.Using $1 = \sin^2 x + \cos^2 x$ and $\sin 2x = 2 \sin x \cos x$, we have $1 + \sin 2x = (\cos x + \sin x)^2$ and $1 - \sin 2x = (\cos x - \sin x)^2$.Thus, $y = 	an^{-1} \sqrt{\frac{(\cos x + \sin x)^2}{(\cos x - \sin x)^2}} = 	an^{-1} \left| \frac{\cos x + \sin x}{\cos x - \sin x} ight|$.Dividing numerator and denominator by $\cos x$, we get $y = 	an^{-1} \left| \frac{1 + 	an x}{1 - 	an x} ight| = 	an^{-1} |	an(\frac{\pi}{4} + x)|$.For $x = \frac{\pi}{6}$, $\frac{\pi}{4} + \frac{\pi}{6} = \frac{5\pi}{12}$, which is in the first quadrant, so $y = \frac{\pi}{4} + x$.Differentiating with respect to $x$, $\frac{dy}{dx} = \frac{d}{dx}(\frac{\pi}{4} + x) = 1$.

If $y = \tan^{-1} \sqrt{\frac{1 + \sin 2x}{1 - \sin 2x}}$, then at $x = \frac{\pi}{6}$, the value of $\frac{dy}{dx}$ is

Similar Questions

Find the derivative of the following function: $\frac{a}{x^{4}}-\frac{b}{x^{2}}+\cos x$,where $a$ and $b$ are fixed non-zero constants.

Find the derivative of the function given by $f(x)=\sin(x^{2}).$

$\frac{d}{dx}(\sin(x^2) + \cos(x^2)) = $ . . . . . .

Suppose $f(x) = \frac{(2^x + 2^{-x}) \tan x \sqrt{\tan^{-1}(x^2 - x + 1)}}{(7x^2 + 3x + 1)^3}$. Then the value of $f'(0)$ is equal to

If $f(1) = 1$ and $f'(1) = 3$,then the derivative of $f(f(f(x))) + (f(x))^2$ at $x = 1$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module