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

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(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 $\tan \theta =$

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