If tan (cos ^{ - 1}x) = sin (cot ^{ - 1}frac{ | vedclass

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

Question

(B) Given that $	an (\cos ^{ - 1}x) = \sin (\cot ^{ - 1}\frac{1}{2})$.Let $\cot ^{ - 1}\frac{1}{2} = \phi$,then $\cot \phi = \frac{1}{2}$.We know tha

Vedclass · Accepted Answer

$\pm \frac{\sqrt{5}}{3}$

Vedclass · Answer

(B) Given that $	an (\cos ^{ - 1}x) = \sin (\cot ^{ - 1}\frac{1}{2})$.Let $\cot ^{ - 1}\frac{1}{2} = \phi$,then $\cot \phi = \frac{1}{2}$.We know that $\sin \phi = \frac{1}{\sqrt{1 + \cot ^2 \phi}} = \frac{1}{\sqrt{1 + (1/2)^2}} = \frac{1}{\sqrt{5/4}} = \frac{2}{\sqrt{5}}$.Let $\cos ^{ - 1}x = 	heta$,then $\cos 	heta = x$.We know that $	an 	heta = \frac{\sqrt{1 - \cos ^2 	heta}}{\cos 	heta} = \frac{\sqrt{1 - x^2}}{x}$.Substituting these into the original equation:$\frac{\sqrt{1 - x^2}}{x} = \frac{2}{\sqrt{5}}$.Squaring both sides:$\frac{1 - x^2}{x^2} = \frac{4}{5}$.$5(1 - x^2) = 4x^2$.$5 - 5x^2 = 4x^2$.$9x^2 = 5$.$x^2 = \frac{5}{9}$.$x = \pm \frac{\sqrt{5}}{3}$.

If $\tan (\cos ^{ - 1}x) = \sin (\cot ^{ - 1}\frac{1}{2})$,then $x =$

Similar Questions

If $\cot \left(\cos^{-1} x\right) = \sec \left(\tan^{-1} \left(\frac{a}{\sqrt{b^2-a^2}}\right)\right)$,where $b > a > 0$,then $x =$

$\tan \left(\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right) = $

The possible values of $x$, which satisfy the trigonometric equation $\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}$ are

The value of $x$, where $x>0$ and $\tan \left(\sec ^{-1}\left(\frac{1}{x}\right)\right)=\sin \left(\tan ^{-1} 2\right)$ is

Considering only the principal values of the inverse trigonometric functions,the set $\{x \geq 0 : \tan^{-1}(2x) + \tan^{-1}(3x) = \frac{\pi}{4}\}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module