cos ^{-1}(cos (-5))+sin ^{-1}(sin (6))-tan ^{ | vedclass

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

Question

(C) We need to evaluate the expression $\cos ^{-1}(\cos (-5))+\sin ^{-1}(\sin (6))-	an ^{-1}(	an (12))$.$1$. For $\cos ^{-1}(\cos (-5))$: Since $\co

Vedclass · Accepted Answer

$4 \pi-11$

Vedclass · Answer

(C) We need to evaluate the expression $\cos ^{-1}(\cos (-5))+\sin ^{-1}(\sin (6))-	an ^{-1}(	an (12))$.$1$. For $\cos ^{-1}(\cos (-5))$: Since $\cos(-x) = \cos(x)$,this is $\cos ^{-1}(\cos (5))$. Since $5 \in [\pi, 2\pi]$,we use $\cos ^{-1}(\cos x) = 2\pi - x$. Thus,$\cos ^{-1}(\cos (5)) = 2\pi - 5$.$2$. For $\sin ^{-1}(\sin (6))$: Since $6 \in [\frac{3\pi}{2}, 2\pi]$,we use $\sin ^{-1}(\sin x) = x - 2\pi$. Thus,$\sin ^{-1}(\sin (6)) = 6 - 2\pi$.$3$. For $	an ^{-1}(	an (12))$: Since $12 \in (3\pi + \frac{\pi}{2}, 4\pi + \frac{\pi}{2})$,we use $	an ^{-1}(	an x) = x - 4\pi$. Thus,$	an ^{-1}(	an (12)) = 12 - 4\pi$.Combining these: $(2\pi - 5) + (6 - 2\pi) - (12 - 4\pi) = 2\pi - 5 + 6 - 2\pi - 12 + 4\pi = 4\pi - 11$.

$\cos ^{-1}(\cos (-5))+\sin ^{-1}(\sin (6))-\tan ^{-1}(\tan (12))$ is equal to :
(The inverse trigonometric functions take the principal values)

Similar Questions

Let $f(x) = \cos \left(2 \tan ^{-1} \sin \left(\cot ^{-1} \sqrt{\frac{1-x}{x}}\right)\right)$ for $0 < x < 1$. Then :

If $y = \tan ^{-1}\left(\frac{1}{1+x+x^{2}}\right) + \tan ^{-1}\left(\frac{1}{x^{2}+2x+3}\right) + \tan ^{-1}\left(\frac{1}{x^{2}+5x+7}\right) + \dots + n \text{ terms}$,then $y'(0)$ is

The number of solutions of the equation $2\tan^{-1}(\cos^2 x) = \tan^{-1}(2\csc^2 x)$ in the interval $[0, 5\pi]$ is $m$. Then which of the following is true?

$\lim _{n \rightarrow \infty} \tan \left\{\sum_{r=1}^{n} \tan ^{-1}\left(\frac{1}{1+r+r^{2}}\right)\right\}$ is equal to..........

Solve $2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \csc x)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$\cos ^{-1}(\cos (-5))+\sin ^{-1}(\sin (6))-\tan ^{-1}(\tan (12))$ is equal to :(The inverse trigonometric functions take the principal values)