Considering only the principal values,if tan | vedclass

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(D) Let $\cot ^{ - 1}(1/2) = 	heta$. Then $\cot 	heta = 1/2$. Since $\cot 	heta = 1/2$,we have a right-angled triangle with adjacent side $1$ and o

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$\sqrt{5}/3$

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(D) Let $\cot ^{ - 1}(1/2) = 	heta$. Then $\cot 	heta = 1/2$. Since $\cot 	heta = 1/2$,we have a right-angled triangle with adjacent side $1$ and opposite side $2$. The hypotenuse is $\sqrt{1^2 + 2^2} = \sqrt{5}$. Thus,$\sin 	heta = 2/\sqrt{5}$. Now,the equation is $	an (\cos ^{ - 1}x) = \sin 	heta = 2/\sqrt{5}$. Let $\cos ^{ - 1}x = \phi$,so $\cos \phi = x$. Then $	an \phi = 2/\sqrt{5}$. In a right-angled triangle with $	an \phi = 2/\sqrt{5}$,the opposite side is $2$ and the adjacent side is $\sqrt{5}$. The hypotenuse is $\sqrt{2^2 + (\sqrt{5})^2} = \sqrt{4 + 5} = \sqrt{9} = 3$. Therefore,$\cos \phi = 	ext{adjacent} / 	ext{hypotenuse} = \sqrt{5}/3$. Hence,$x = \sqrt{5}/3$.

Considering only the principal values,if $\tan (\cos ^{ - 1}x) = \sin [\cot ^{ - 1}(1/2)]$,then $x$ is equal to

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