If x takes non-positive permissible value,the | vedclass

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

Question

(B) Let $\sin^{-1} x = y$. Then $x = \sin y$. Since $x$ is non-positive and permissible,we have $-1 \le x \le 0$. This implies $-\frac{\pi}{2} \le \si

Vedclass · Accepted Answer

$-\cos^{-1} \sqrt{1 - x^2}$

Vedclass · Answer

(B) Let $\sin^{-1} x = y$. Then $x = \sin y$. Since $x$ is non-positive and permissible,we have $-1 \le x \le 0$. This implies $-\frac{\pi}{2} \le \sin^{-1} x \le 0$,so $-\frac{\pi}{2} \le y \le 0$. We know that $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$ for $y \in [0, \pi]$. Since $y \in [-\frac{\pi}{2}, 0]$,we have $-y \in [0, \frac{\pi}{2}]$. Thus,$\cos(-y) = \sqrt{1 - x^2}$. Since $\cos(-y) = \cos y$,we have $\cos y = \sqrt{1 - x^2}$. Taking the inverse cosine on both sides for the range $[0, \frac{\pi}{2}]$,we get $-y = \cos^{-1} \sqrt{1 - x^2}$. Therefore,$y = -\cos^{-1} \sqrt{1 - x^2}$. Hence,$\sin^{-1} x = -\cos^{-1} \sqrt{1 - x^2}$.

If $x$ takes non-positive permissible value,then $\sin^{-1} x =$

Similar Questions

$\tan \left[ 2\tan^{-1}\left( \frac{1}{5} \right) - \frac{\pi}{4} \right] = $

If $\cos^{-1}\left(\frac{2}{3x}\right) + \cos^{-1}\left(\frac{3}{4x}\right) = \frac{\pi}{2}$ for $x > \frac{3}{4}$,then $x$ is equal to

If $y = \tan^{-1}\left(\frac{1}{x^2 + x + 1}\right) + \tan^{-1}\left(\frac{1}{x^2 + 3x + 3}\right) + \tan^{-1}\left(\frac{1}{x^2 + 5x + 7}\right) + \dots$ up to $n$ terms,then $\frac{dy}{dx}$ is equal to

If $\sin ^{ - 1}\frac{{2a}}{{1 + {a^2}}} - \cos ^{ - 1}\frac{{1 - {b^2}}}{{1 + {b^2}}} = \tan ^{ - 1}\frac{{2x}}{{1 - {x^2}}}$,then $x = $

If $\cos^{-1} p + \cos^{-1} q + \cos^{-1} r = \pi$,then $p^2 + q^2 + r^2 + 2pqr = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module