If x = {sin ^{ - 1}}(sin 10) and y = {cos ^{ | vedclass

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

Question

(A) We know that the principal value range for $\sin^{-1}(\sin 	heta)$ is $[-\pi/2, \pi/2]$.Since $3\pi \approx 9.42$ and $10$ lies in the interval $

Vedclass · Accepted Answer

$\pi $

Vedclass · Answer

(A) We know that the principal value range for $\sin^{-1}(\sin 	heta)$ is $[-\pi/2, \pi/2]$.Since $3\pi \approx 9.42$ and $10$ lies in the interval $(2\pi, 3\pi)$,we use the property $\sin^{-1}(\sin 	heta) = 	heta - 3\pi$ for $	heta \in [2\pi, 3\pi]$.Thus,$x = \sin^{-1}(\sin 10) = 10 - 3\pi$.However,looking at the graph provided,the value is $3\pi - 10$.Let's re-evaluate: $10$ radians is in the third quadrant ($3\pi Actually,$3\pi Since $3\pi - 10 \in [-\pi/2, \pi/2]$,$x = 3\pi - 10$.For $y = \cos^{-1}(\cos 10)$,the range is $[0, \pi]$.Since $3\pi Thus,$y = 4\pi - 10$.Therefore,$y - x = (4\pi - 10) - (3\pi - 10) = \pi$.

If $x = {\sin ^{ - 1}}(\sin 10)$ and $y = {\cos ^{ - 1}}(\cos 10)$,then $y - x$ is equal to

Similar Questions

The value of $\sin^{-1}(\sin \frac{7\pi}{6}) + \cos^{-1}(\cos \frac{7\pi}{6}) + \tan^{-1}(\tan \frac{2\pi}{3})$ is equal to...

If $\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \pi$,then $x + y + z$ is equal to

$\sin^{-1} \frac{1}{\sqrt{5}} + \cot^{-1} 3$ is equal to

If for $|x|>1$, $\tanh ^{-1}\left(\frac{1}{x}\right)+\operatorname{coth}^{-1}(x)=\log _e(f(x))$, then $f(-5)=$

If $\cos^{-1} \frac{3}{5} - \sin^{-1} \frac{4}{5} = \cos^{-1} x$,then $x = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module