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If $\tan ^{-1}y=\tan ^{-1}x+\tan ^{-1}\left( \dfrac{2x}{1-x^2} \right)$ where $|x| < \dfrac {1}{\sqrt {3}}$ then find the value of $y$.

. . . Let $\tan^{-1} y= A$ $$y=\tan A$$

$$\tan^{-1} x=B$$ $$x=\tan B$$

$$ \tan ^{-1}\left( \dfrac{2x}{1-x^2} \right) =C $$

$$\dfrac {2x}{1-x^2}=\tan C$$.

StubbornAtom
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Aryabhatta
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2 Answers2

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We know that $$tan2A= \frac{2tanA}{1-tan^2A}$$

Now, substitute $tanA=x\implies tan^{-1}x=A$

Plugging this into first equation, $tan2(tan^{-1}x)= \frac{2x}{1-x^2}$

Again taking $tan^{-1}$ on both sides we get

$$2tan^{-1}x = tan^{-1}\frac{2x}{1-x^2}$$

Plugging this into the question,

$$tan^{-1}y= tan^{-1}x + 2tan^{-1}x = 3tan^{-1}x$$

We know that $$tan3A=tan(2A+A)= \frac{tanA+tan2A}{1-tanAtan2A}$$

This can be reduced to $$tan3A=\frac{3tanA-tan^3A}{1-3tan^2A}$$

By a similiar logic as before, $$3tan^{-1}x= tan^{-1}\frac{3x-x^3}{1-3x^2}$$

Now the question reduces to $$tan^{-1}y=tan^{-1}\frac{3x-x^2}{1-3x^2}$$

So, we conclude that $$y=\frac{3x-x^3}{1-3x^2}$$

NOTE: Since it is given that $|x|\lt\frac{1}{\sqrt3}$ we can safely apply the above formulae without adding or subtracting $\pi$,because x is well within the domain.

Ananth Kamath
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We use the identity $$\tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}.$$

Letting $a=\tan^{-1}(x)$ and $b=\tan^{-1}\left(\frac{2x}{1-x^2}\right)$, we have $$y=\tan(a+b)=\frac{x+\frac{2x}{1-x^2}}{1-\frac{2x^2}{1-x^2}}=\frac{3x-x^3}{1-3x^2}$$

helloworld112358
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