If sin ^{-1} x+sin ^{-1} y=frac{pi}{3} and co | vedclass

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(C) Given that $\cot ^{-1}\left(\frac{1}{x}\right)-\cot ^{-1}\left(\frac{1}{y}\right)=0$,which implies $\cot ^{-1}\left(\frac{1}{x}\right)=\cot ^{-1}\

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$\frac{1}{2}$

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(C) Given that $\cot ^{-1}\left(\frac{1}{x}\right)-\cot ^{-1}\left(\frac{1}{y}\right)=0$,which implies $\cot ^{-1}\left(\frac{1}{x}\right)=\cot ^{-1}\left(\frac{1}{y}\right)$.This means $\frac{1}{x}=\frac{1}{y}$,so $x=y$.Substitute $x=y$ into the first equation: $\sin ^{-1} x+\sin ^{-1} x=\frac{\pi}{3}$.$2 \sin ^{-1} x=\frac{\pi}{3} \implies \sin ^{-1} x=\frac{\pi}{6}$.Therefore,$x=\sin \left(\frac{\pi}{6}\right)=\frac{1}{2}$.Since $x=y$,we have $y=\frac{1}{2}$.Now,calculate $2 x^2+y^2-x y$:$2 \left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = 2 \left(\frac{1}{4}\right)+\frac{1}{4}-\frac{1}{4} = \frac{1}{2}+0 = \frac{1}{2}$.

If $\sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{3}$ and $\cot ^{-1}\left(\frac{1}{x}\right)-\cot ^{-1}\left(\frac{1}{y}\right)=0$,then $2 x^2+y^2-x y=$

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