If {cos ^{ - 1}}left( {frac{1}{x}} right) = t | vedclass

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

Question

(D) Given that ${\cos ^{ - 1}}\left( {\frac{1}{x}} ight) = 	heta $. This implies $\cos 	heta = \frac{1}{x}$. We know that $\sin^2 	heta + \cos^2

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Given that ${\cos ^{ - 1}}\left( {\frac{1}{x}} ight) = 	heta $. This implies $\cos 	heta = \frac{1}{x}$. We know that $\sin^2 	heta + \cos^2 	heta = 1$,so $\sin 	heta = \sqrt{1 - \cos^2 	heta} = \sqrt{1 - \left(\frac{1}{x}ight)^2} = \sqrt{\frac{x^2 - 1}{x^2}} = \frac{\sqrt{x^2 - 1}}{x}$. Now,$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{\frac{\sqrt{x^2 - 1}}{x}}{\frac{1}{x}} = \sqrt{x^2 - 1}$. Therefore,the correct option is $D$.

If ${\cos ^{ - 1}}\left( {\frac{1}{x}} \right) = \theta $,then $\tan \theta =$

Similar Questions

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

$\sec^{-1}[\sec(-30^o)] = $ ....... $^o$

$\cos \left[\cot ^{-1}(-\sqrt{3})+\frac{\pi}{6}\right]$ is equal to

Find the value of $\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)$.

Let the function $g: (-\infty, \infty) \rightarrow \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ be given by $g(u) = 2 \tan^{-1}(e^u) - \frac{\pi}{2}$. Then,$g$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module