Find the value of tan ^{-1}(1)+cos ^{-1}left( | vedclass

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

Question

(A) Let $	an ^{-1}(1)=x$. Then,$	an x=1=	an \left(\frac{\pi}{4}ight)$. Thus,$	an ^{-1}(1)=\frac{\pi}{4}$.Let $\cos ^{-1}\left(-\frac{1}{2}ight

Vedclass · Accepted Answer

$\frac{3 \pi}{4}$

Vedclass · Answer

(A) Let $	an ^{-1}(1)=x$. Then,$	an x=1=	an \left(\frac{\pi}{4}ight)$. Thus,$	an ^{-1}(1)=\frac{\pi}{4}$.Let $\cos ^{-1}\left(-\frac{1}{2}ight)=y$. Then,$\cos y=-\frac{1}{2}=-\cos \left(\frac{\pi}{3}ight)=\cos \left(\pi-\frac{\pi}{3}ight)=\cos \left(\frac{2 \pi}{3}ight)$. Thus,$\cos ^{-1}\left(-\frac{1}{2}ight)=\frac{2 \pi}{3}$.Let $\sin ^{-1}\left(-\frac{1}{2}ight)=z$. Then,$\sin z=-\frac{1}{2}=-\sin \left(\frac{\pi}{6}ight)=\sin \left(-\frac{\pi}{6}ight)$. Thus,$\sin ^{-1}\left(-\frac{1}{2}ight)=-\frac{\pi}{6}$.Substituting these values,we get:$	an ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}ight)+\sin ^{-1}\left(-\frac{1}{2}ight) = \frac{\pi}{4}+\frac{2 \pi}{3}-\frac{\pi}{6}$.Taking the least common multiple of the denominators $4, 3, 6$,which is $12$:$= \frac{3 \pi + 8 \pi - 2 \pi}{12} = \frac{9 \pi}{12} = \frac{3 \pi}{4}$.

Find the value of $\tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)$

Similar Questions

If $\tan \theta = - \frac{1}{\sqrt{3}}$,$\sin \theta = \frac{1}{2}$,and $\cos \theta = - \frac{\sqrt{3}}{2}$,then the principal value of $\theta$ is:

If ${\cot ^{ - 1}}x + {\tan ^{ - 1}}3 = \frac{\pi }{2}$,then $x =$

Write the function in the simplest form: $\tan ^{-1} \frac{x}{\sqrt{a^{2}-x^{2}}}, |x| < a$

If $y = \sin^{-1} \left( \frac{2x}{1 + x^2} \right)$,where $0 < x < 1$ and $0 < y < \frac{\pi}{2}$,then $\frac{dy}{dx} = $

$\tan ^{-1}(-\sqrt{3})-\sec ^{-1}(-2)=$ . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module