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

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(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

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

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(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 $x =$

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