sin^{-1} frac{1}{sqrt{5}} + cot^{-1} 3 is equ | vedclass

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(B) Let $\alpha = \sin^{-1} \frac{1}{\sqrt{5}}$. Then $\sin \alpha = \frac{1}{\sqrt{5}}$.Using the identity $\sin^2 \alpha + \cos^2 \alpha = 1$,we hav

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$\frac{\pi}{4}$

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(B) Let $\alpha = \sin^{-1} \frac{1}{\sqrt{5}}$. Then $\sin \alpha = \frac{1}{\sqrt{5}}$.Using the identity $\sin^2 \alpha + \cos^2 \alpha = 1$,we have $\cos \alpha = \sqrt{1 - \frac{1}{5}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$.Thus,$	an \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{1/\sqrt{5}}{2/\sqrt{5}} = \frac{1}{2}$.Therefore,$\alpha = 	an^{-1} \frac{1}{2} = \cot^{-1} 2$.Now,the expression becomes $\cot^{-1} 2 + \cot^{-1} 3$.Using the formula $\cot^{-1} x + \cot^{-1} y = \cot^{-1} \left( \frac{xy - 1}{x + y} ight)$ for $x, y > 0$:$\cot^{-1} 2 + \cot^{-1} 3 = \cot^{-1} \left( \frac{2 	imes 3 - 1}{2 + 3} ight) = \cot^{-1} \left( \frac{5}{5} ight) = \cot^{-1} (1) = \frac{\pi}{4}$.

$\sin^{-1} \frac{1}{\sqrt{5}} + \cot^{-1} 3$ is equal to

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