Suppose tan ^{-1} y = tan ^{-1} x + tan ^{-1} | vedclass

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(D) Given the equation: $	an ^{-1} y = 	an ^{-1} x + 	an ^{-1} \left( \frac{2x}{1 - x^2} ight)$.Since $|x| < \frac{1}{\sqrt{3}}$,the condition fo

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$\frac{3x - x^3}{1 - 3x^2}$

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(D) Given the equation: $	an ^{-1} y = 	an ^{-1} x + 	an ^{-1} \left( \frac{2x}{1 - x^2} ight)$.Since $|x| Substituting this into the equation,we get:$	an ^{-1} y = 	an ^{-1} x + 2 	an ^{-1} x$$	an ^{-1} y = 3 	an ^{-1} x$Using the identity $3 	an ^{-1} x = 	an ^{-1} \left( \frac{3x - x^3}{1 - 3x^2} ight)$ for $|x| $	an ^{-1} y = 	an ^{-1} \left( \frac{3x - x^3}{1 - 3x^2} ight)$Therefore,$y = \frac{3x - x^3}{1 - 3x^2}$.

Suppose $\tan ^{-1} y = \tan ^{-1} x + \tan ^{-1} \left( \frac{2x}{1 - x^2} \right)$,where $|x| < \frac{1}{\sqrt{3}}$,then one of the values of $y$ is

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