A Problem on Inverse Trigonometric Functions and Differential Equation. $f(x)=\left(\sin \left(\tan ^{-1} x\right)+\sin \left(\cot ^{-1} x\right)\right)^{2}-1,\ |x|&gt;1$
If $\displaystyle\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} x}\left(\sin ^{-1}(f(x))\right)$ and
$y(\sqrt{3})=\frac{\pi}{6}$, then $y(-\sqrt{3})$ is equal to :
Options:
$1. \quad-\frac{\pi}{6}\\
2. \qquad \frac{2 \pi}{3}\\
3. \qquad \frac{5 \pi}{6}\\
4. \qquad\frac{\pi}{3}$

It can be easily shown that $f(x) = [\sin(\tan^{-1}x) + \sin(\cot^{-1}x  )]^2 -1 = \frac{2x}{1+x^2}$.
$\frac{d}{dx} (\sin^{-1}\frac{2x}{1+x^2}) = \frac{-2}{1+x^2}$
Now $\frac{dy}{dx} = \frac{1}{2} \frac{ d(\sin^{-1}f(x))}{ dx} = \frac{-1}{1+x^2}$ from which we get $y = - \tan^{-1}x  + C. $
Now $y (\sqrt 3) = \frac{\pi}{6}$. So $C= \frac{\pi}{2}$.
So $y = \frac{\pi}{2} -  \tan^{-1}x $.
$y(-\sqrt 3) = \frac{5\pi}{6}$
Have I gone wrong any where? Can anyone please check my solution?
$$\dfrac{d}{dx}(\sin^{-1}f)= \dfrac{1}{\sqrt{1-f^2}}\cdot \dfrac{df}{dx}=P\cdot Q \;;$$
$$P=\dfrac{1}{\sqrt{1-(2x/(1+x^2))^2}} = \dfrac{1+x^2}{1-x^2} $$
Differentiating by Quotient rule
$$Q=\dfrac{2}{(1+x)^2}; \quad PQ=\dfrac{2}{1-x^2};\; $$
EDIT1:
Stopped too early by mistake.  Continuing,
$$\dfrac{dy}{dx}= \dfrac{1}{1-x^2}$$
Integrates to
$$y=\log_e\sqrt {\dfrac{1+x}{1-x}} +c$$
given
$$ y(\sqrt3)=\dfrac{\pi}{6}$$
$y$ is not real, constant of integration cannot be evaluated. For this result no option available.
$$\dfrac{d}{dx}(\sin^{-1}f)= \dfrac{1}{\sqrt{1-f^2}}\cdot \dfrac{df}{dx}=P\cdot Q \;;$$
$$P=\dfrac{1}{\sqrt{1-(2x/(1+x^2))^2}} = \dfrac{1+x^2}{1-x^2} $$
Differentiating by Quotient rule
$$Q=\dfrac{2}{(1+x)^2}; \quad PQ=\dfrac{2}{1-x^2};\; $$
EDIT1:
Stopped too early by mistake.  Continuing,
$$\dfrac{dy}{dx}= \dfrac{1}{1-x^2}$$
Integrates to
$$y=\log_e\sqrt {\dfrac{1+x}{1-x}} +c$$
given
$$ y(\sqrt3)=\dfrac{\pi}{6}$$
$y$ is not real, constant of integration cannot be evaluated. For this result no option available.