The solution of tan ^{-1} x+2 cot ^{-1} x=fra | vedclass

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(D) Given equation: $	an ^{-1} x+2 \cot ^{-1} x=\frac{2 \pi}{3}$ We know that $\cot ^{-1} x = 	an ^{-1} (\frac{1}{x})$ for $x > 0$. Substituting thi

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$\sqrt{3}$

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(D) Given equation: $	an ^{-1} x+2 \cot ^{-1} x=\frac{2 \pi}{3}$ We know that $\cot ^{-1} x = 	an ^{-1} (\frac{1}{x})$ for $x > 0$. Substituting this,we get: $	an ^{-1} x + 2 	an ^{-1} (\frac{1}{x}) = \frac{2 \pi}{3}$ Using the identity $2 	an ^{-1} y = 	an ^{-1} (\frac{2y}{1-y^2})$,we have: $	an ^{-1} x + 	an ^{-1} (\frac{2/x}{1-1/x^2}) = \frac{2 \pi}{3}$ $	an ^{-1} x + 	an ^{-1} (\frac{2x}{x^2-1}) = \frac{2 \pi}{3}$ Applying $	an ^{-1} A + 	an ^{-1} B = 	an ^{-1} (\frac{A+B}{1-AB})$: $	an ^{-1} (\frac{x + \frac{2x}{x^2-1}}{1 - x(\frac{2x}{x^2-1})}) = \frac{2 \pi}{3}$ $\frac{\frac{x^3-x+2x}{x^2-1}}{\frac{x^2-1-2x^2}{x^2-1}} = 	an (\frac{2 \pi}{3})$ $\frac{x^3+x}{-x^2-1} = -\sqrt{3}$ $\frac{x(x^2+1)}{-(x^2+1)} = -\sqrt{3}$ $-x = -\sqrt{3} \implies x = \sqrt{3}$

The solution of $\tan ^{-1} x+2 \cot ^{-1} x=\frac{2 \pi}{3}$ is

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