If cos^{-1} x = alpha (0

Question

(B) Given equation is $\sin^{-1} (2 x \sqrt{1 - x^2}) + \sec^{-1} (\frac{1}{2 x^2 - 1}) = \frac{2 \pi}{3}$.Since $\cos^{-1} x = \alpha$,we have $x = \

Vedclass · Accepted Answer

$\frac{\pi}{6}$

Vedclass · Answer

(B) Given equation is $\sin^{-1} (2 x \sqrt{1 - x^2}) + \sec^{-1} (\frac{1}{2 x^2 - 1}) = \frac{2 \pi}{3}$.Since $\cos^{-1} x = \alpha$,we have $x = \cos \alpha$. Given $0 Substituting $x = \cos \alpha$ into the equation:$\sin^{-1} (2 \cos \alpha \sqrt{1 - \cos^2 \alpha}) + \sec^{-1} (\frac{1}{2 \cos^2 \alpha - 1}) = \frac{2 \pi}{3}$$\sin^{-1} (2 \cos \alpha \sin \alpha) + \sec^{-1} (\frac{1}{\cos 2 \alpha}) = \frac{2 \pi}{3}$$\sin^{-1} (\sin 2 \alpha) + \cos^{-1} (\cos 2 \alpha) = \frac{2 \pi}{3}$Since $0 Therefore,$2 \alpha + 2 \alpha = \frac{2 \pi}{3}$$4 \alpha = \frac{2 \pi}{3}$$\alpha = \frac{\pi}{6}$.

If $\cos^{-1} x = \alpha$ $(0 < x < 1)$ and $\sin^{-1} (2 x \sqrt{1 - x^2}) + \sec^{-1} (\frac{1}{2 x^2 - 1}) = \frac{2 \pi}{3}$,then $\alpha$ is

Similar Questions

If $y = \sin^{-1} \left( \frac{5x + 12\sqrt{1-x^2}}{13} \right)$,then $\frac{dy}{dx} = $

The equation $2\cos^{-1}x + \sin^{-1}x = \frac{11\pi}{6}$ has

If $f(x) = 2 \tan^{-1} x + \sin^{-1} \left( \frac{2x}{1 + x^2} \right)$,where $x > 1$,then $f(5)$ is equal to:

$\sin^{-1} x + \cos^{-1} x$ is equal to

Let $S$ be the set of all solutions of the equation $\cos ^{-1}(2 x)-2 \cos ^{-1}\left(\sqrt{1-x^2}\right)=\pi$,where $x \in\left[-\frac{1}{2}, \frac{1}{2}\right]$. Then $\sum_{x \in S} 2 \sin ^{-1}\left(x^2-1\right)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module