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Find the values of $x$ such that $$2\tan^{-1}x+\sin^{-1}\left(\frac{2x}{1+x^2}\right)$$ is independent of $x$.

Checking for $x\in [-1,1]$ In the taken domain $\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ comes out to be $2\tan^{-1}x$ hence the taken function comes out to be equal to $4\tan^{-1}x$ hence the function is clearly dependent on $x$.

Now checking for $x\in (1,\infty)$ In the taken domain $2\tan^{-1}x$ comes out to be $\pi-\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ and hence the net sum becomes independent of $x$.

Now checking for $x\in (-\infty,-1)$ In the taken domain $2\tan^{-1}x$ comes out to be $-\pi-\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ and hence the net sum becomes $-\pi$ therefore becomes, independent of $x$.

But the answer has been mentioned as just $x\in [1,\infty)$ Can anybody tell me why the second set has not been included.

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    $\begingroup$ Good question. I think that either whoever gives the question didn't take into account the negative option or the question is meant to be solved just for $\;x>0\;$ . I get the same result as you with a little calculus. $\endgroup$ – DonAntonio Jun 24 '16 at 20:24
  • $\begingroup$ @Joanpemo Can you also show me the calculus method? $\endgroup$ – Harsh Sharma Jun 24 '16 at 20:27
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    $\begingroup$ Wrote it dwon in an answer as it is a little long. $\endgroup$ – DonAntonio Jun 24 '16 at 20:28
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The author of the question may have been thinking purely geometrically, picturing the situation below, with $x$ the (non-negative) length of a segment:

enter image description here

For completeness, here's the argument one can make ...

Writing $\alpha := \angle BAC = \angle BAD$ and $\beta := \angle CBD$, we have $$\tan \alpha = x \qquad\text{and}\qquad \sin\beta = \frac{2x}{1+x^2} = \sin 2\alpha$$ (the latter by an aspect of the Law of Sines).

Certainly, $\alpha = \operatorname{atan}{x}$ for all $x$. On the other hand, $\operatorname{asin}\left(\frac{2x}{1+x^2}\right)$ is either $\beta$ or $2\alpha$, depending upon which is non-obtuse; this condition depends upon how $\alpha$ compares to $\pi/4$, which in turn depends upon how $x$ compares to $1$. Therefore, noting that opposite angles in a cyclic quadrilateral are supplementary, we can write:

$$2 \operatorname{atan} x + \operatorname{asin}\left(\frac{2x}{1+x^2}\right) \;=\; \begin{cases} 2\alpha + \phantom{2}\beta = \pi &\text{, for } x \geq 1 \\[4pt] 2\alpha + 2\alpha = 4 \operatorname{atan}x &\text{, otherwise} \end{cases}$$


Without an explicit restriction on the sign of $x$, though, you're correct: the solution to the problem stated should be $|x| \geq 1$, including negative values of $x$.

Edit. Strictly speaking, the answer cannot be $|x| \geq 1$, because then the sign of the sum in question would depend upon the sign of $x$. Perhaps the most accurate way to describe the result is:

  • The sum is constant (with value $\pi$) for $x \geq 1$.
  • The sum is constant (with value $-\pi$) for $x \leq -1$.
  • The sum is non-constant (with value $4\operatorname{atan}x$) for $-1< x < 1$.
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    $\begingroup$ Thanks a lot for a whole new explanatory point of view. $\endgroup$ – Harsh Sharma Jun 25 '16 at 4:05
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Differentiate it:

$$\left(2\arctan x+\arcsin\frac{2x}{1+x^2}\right)'=\frac2{1+x^2}+\frac{2-2x^2}{(1+x^2)^2}\frac1{\sqrt{1-\frac{4x^2}{(1+x^2)^2}}}=$$

$$=\frac2{1+x^2}+\frac{2-2x^2}{(1+x^2)^2}\frac{1+x^2}{\sqrt{(1-x^2)^2}}=\frac2{1+x^2}+\frac{2(1-x^2)}{(1+x^2)|1-x^2|}$$

We can thus see that if $\;|1-x^2|=-(1-x^2)\iff 1-x^2<0\iff |x|>1\;$ , then the above last expression is zero and thus the function is a constant one and doesn't depend on $\;x\;$ .

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    $\begingroup$ Thanks a lot for such a wonderful method. $\endgroup$ – Harsh Sharma Jun 25 '16 at 4:04
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Let $\tan^{-1}x=y\implies-\dfrac\pi2\le y\le\dfrac\pi2\iff-\pi\le2y\le\pi$

$$\implies\dfrac{2x}{1+x^2}=\sin2y$$

$$\sin^{-1}\dfrac{2x}{1+x^2}=\begin{cases}-2y-\pi &\mbox{if }2y<-\dfrac\pi2\iff x<-1 \\ 2y & \mbox{if }-\dfrac\pi2\le2y\le\dfrac\pi2\\ \pi-2y & \mbox{if }2y>\dfrac\pi2\end{cases}$$

Can you take it home from here?

See also: Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$

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  • $\begingroup$ Thank you sir. I followed the same procedure in my own attempt too and hence was confused in the $-\pi-2y$ part. $\endgroup$ – Harsh Sharma Jun 25 '16 at 5:06

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