Given that the inverse trigonometric function | vedclass

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

Question

(B) We are given $\cos ^{-1} x - \sin ^{-1} y = \alpha$. Using the identity $\sin ^{-1} y = \frac{\pi}{2} - \cos ^{-1} y$,we get: $\cos ^{-1} x - (\fr

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) We are given $\cos ^{-1} x - \sin ^{-1} y = \alpha$. Using the identity $\sin ^{-1} y = \frac{\pi}{2} - \cos ^{-1} y$,we get: $\cos ^{-1} x - (\frac{\pi}{2} - \cos ^{-1} y) = \alpha$ $\cos ^{-1} x + \cos ^{-1} y = \frac{\pi}{2} + \alpha$. Since $\alpha \in [-\frac{\pi}{2}, \pi]$,we have $\frac{\pi}{2} + \alpha \in [0, \frac{3\pi}{2}]$. Taking cosine on both sides: $\cos(\cos ^{-1} x + \cos ^{-1} y) = \cos(\frac{\pi}{2} + \alpha)$ $xy - \sqrt{1-x^2}\sqrt{1-y^2} = -\sin \alpha$ $xy + \sin \alpha = \sqrt{1-x^2}\sqrt{1-y^2}$. Squaring both sides: $(xy + \sin \alpha)^2 = (1-x^2)(1-y^2)$ $x^2y^2 + 2xy \sin \alpha + \sin^2 \alpha = 1 - x^2 - y^2 + x^2y^2$ $x^2 + y^2 + 2xy \sin \alpha = 1 - \sin^2 \alpha$ $x^2 + y^2 + 2xy \sin \alpha = \cos^2 \alpha$. The minimum value of $\cos^2 \alpha$ is $0$,which occurs when $\alpha = \frac{\pi}{2}$. Thus,the minimum value is $0$.

Given that the inverse trigonometric function assumes principal values only. Let $x, y$ be any two real numbers in $[-1, 1]$ such that $\cos ^{-1} x - \sin ^{-1} y = \alpha$,where $-\frac{\pi}{2} \leq \alpha \leq \pi$. Then,the minimum value of $x^2 + y^2 + 2xy \sin \alpha$ is

Similar Questions

$\tan \left(\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right) = $

If $y = \cos \left(\frac{\pi}{3} + \cos^{-1} \frac{x}{2}\right)$,then $(x - y)^2 + 3y^2$ is equal to . . . . . . .

Find $\frac{dx}{dy}$ if $y = \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right)$ for $-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$.

If $\sin ^{-1} x-\cos ^{-1} 2 x=\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)-\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$,then $\tan ^{-1} x+\tan ^{-1}\left(\frac{x}{x+1}\right)=$

The value of $x$,where $x>0$ and $\tan \left(\sec ^{-1}\left(\frac{1}{x}\right)\right)=\sin \left(\tan ^{-1} 2\right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module