If {sin ^{ - 1}}x + {cot ^{ - 1}}left( {frac{ | vedclass

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(B) Given equation: ${\sin ^{ - 1}}x + {\cot ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{2}$.We know that ${\sin ^{ - 1}}x + {\cos ^{ - 1}}x =

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$\frac{1}{{\sqrt 5 }}$

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(B) Given equation: ${\sin ^{ - 1}}x + {\cot ^{ - 1}}\left( {\frac{1}{2}} ight) = \frac{\pi }{2}$.We know that ${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \frac{\pi }{2}$.Also,${\cot ^{ - 1}}\left( {\frac{1}{2}} ight) = {	an ^{ - 1}}(2)$.Let ${	an ^{ - 1}}(2) = 	heta$,then $	an 	heta = 2$. Using the triangle method,if the opposite side is $2$ and the adjacent side is $1$,the hypotenuse is $\sqrt{2^2 + 1^2} = \sqrt{5}$.Thus,$\cos 	heta = \frac{1}{\sqrt{5}}$,which implies $	heta = {\cos ^{ - 1}}\left( {\frac{1}{\sqrt{5}}} ight)$.Substituting this into the original equation: ${\sin ^{ - 1}}x + {\cos ^{ - 1}}\left( {\frac{1}{\sqrt{5}}} ight) = \frac{\pi }{2}$.Comparing this with ${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \frac{\pi }{2}$,we get $x = \frac{1}{\sqrt{5}}$.

If ${\sin ^{ - 1}}x + {\cot ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{2},$ then $x$ is

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