The value of tan ^{-1}(-1)+sec ^{-1}(-2)+sin | vedclass

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

Question

(D) We know the principal value branches of the inverse trigonometric functions:$1$. $	an ^{-1}(-x) = -	an ^{-1}(x)$,so $	an ^{-1}(-1) = -\frac{\pi

Vedclass · Accepted Answer

$\frac{2 \pi}{3}$

Vedclass · Answer

(D) We know the principal value branches of the inverse trigonometric functions:$1$. $	an ^{-1}(-x) = -	an ^{-1}(x)$,so $	an ^{-1}(-1) = -\frac{\pi}{4}$.$2$. $\sec ^{-1}(-x) = \pi - \sec ^{-1}(x)$,so $\sec ^{-1}(-2) = \pi - \sec ^{-1}(2) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.$3$. $\sin ^{-1}(\frac{1}{\sqrt{2}}) = \frac{\pi}{4}$.Adding these values together:$-\frac{\pi}{4} + \frac{2\pi}{3} + \frac{\pi}{4} = \frac{2\pi}{3}$.

The value of $\tan ^{-1}(-1)+\sec ^{-1}(-2)+\sin ^{-1} \frac{1}{\sqrt{2}}$ is . . . . . . .

Similar Questions

The value of $\cos (\tan^{-1} x)$ is equal to : . . . . . . (Where $|x| < 1$)

$e^{\log (\cosh^{-1} 2)}$ is equal to :

$\sin (2{\sin ^{ - 1}}0.8) = $

The value of $\sin ^{-1}\left(\frac{-1}{2}\right)+\sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)$ is

The value of $x$ which satisfies the equation $\tan^{-1}x = \sin^{-1}\left(\frac{3}{\sqrt{10}}\right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module