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Solve

$$\tan^{-1}\left(\frac{1-x}{1+x}\right)=\frac12\tan^{-1}(x)$$

I used the formula

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

and got the final answer as

$$\tan^{-1}\left(\frac{1-x}{1+x}\right)=\frac12\tan^{-1}\left(\frac{1-x^2}{2(x)}\right)$$

which is not right.

Community
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m0bi5
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4 Answers4

2

Let $\theta = \tan^{-1}\left(\dfrac{1-x}{1+x}\right) \to \tan \theta = \dfrac{1-x}{1+x} \to 2\theta = \tan^{-1}x \to x = \tan(2\theta) = \dfrac{2\tan \theta}{1-\tan^2(\theta)} = \dfrac{2\left(\dfrac{1-x}{1+x}\right)}{1-\left(\dfrac{1-x}{1+x}\right)^2}$. Can you continue?

DeepSea
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$$\arctan \frac{1-x}{1+x}=\frac12\arctan x\implies \arctan\tan(\pi/4-\arctan(x))=\frac12\arctan x$$ So: $$\pi/4=\frac32\arctan x\implies \arctan x=\frac{\pi}{6}\implies x=\tan(\pi/6)=1/\sqrt3$$

RE60K
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  • $$\arctan\left[\tan\left(\dfrac\pi4-\arctan x\right)\right]=\dfrac\pi4-\arctan x$$ $$\iff-\dfrac\pi2\le\dfrac\pi4-\arctan x\le\dfrac\pi2\iff\dfrac\pi2\ge\arctan x-\dfrac\pi4\ge-\dfrac\pi2$$ $$\iff\dfrac{3\pi}4\ge\arctan x\ge-\dfrac\pi4\iff x\ge-1$$ – lab bhattacharjee Mar 07 '15 at 16:50
  • @labbhattacharjee can't get you. – RE60K Mar 08 '15 at 06:23
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Using the tangent addition formula, $$ \tan^{-1} a + \tan^{-1}b = \tan^{-1} \frac{a+b}{1-ab}$$ let $a = 1$ and $b = -x$, so $$ \frac{\pi}{4} - \tan^{-1} x = \tan^{-1} \frac{1-x}{1+x} = \frac{1}{2}\tan^{-1} x$$ So then $ \frac{\pi}{6} = \tan^{-1} x$ and $x = 1/\sqrt 3$.

Victor Liu
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1

Like my answer here,

$$2\tan^{-1}\frac{1-x}{1+x}=\tan^{-1}\dfrac{2\cdot\dfrac{1-x}{1+x}}{1-\left(\dfrac{1-x}{1+x}\right)^2}$$ if $\left(\dfrac{1-x}{1+x}\right)^2<1\iff\left(\dfrac{1-x}{1+x}\right)^2-1<0\iff-\dfrac{4x}{(1+x)^2}<0\iff x>0$

In that case, $$\tan^{-1}x=\tan^{-1}\dfrac{2\cdot\dfrac{1-x}{1+x}}{1-\left(\dfrac{1-x}{1+x}\right)^2}=\tan^{-1}\dfrac{1-x^2}{2x}$$

$$\iff x=\dfrac{1-x^2}{2x}\iff x^2=\dfrac13$$

But we need $x>0$

Community
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lab bhattacharjee
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