Express $\arcsin(x)$ in terms of $\arccos(x)$.
Using the same, solve the equation
$$ 2\,\tan^{-1}x = \sin^{-1} x + \cos^{-1} x $$
I'm not sure if I am on the right track, but here is what i did:
$$\sin\left(\frac{\pi}{2}-x\right) = \cos(x)$$
$$\sin(x) = \frac{\pi}{2}-\cos(x)$$
Best Answer
$$x = \sin(y)$$ $$x = \cos\left(\frac{\pi}{2}-y\right)$$
$$y=\arcsin(x)$$ $$\frac{\pi}{2}-y=\arccos(x)$$
Adding the last equations will give your identity: $$\frac{\pi}{2}=\arcsin(x)+\arccos(x)$$
Now you can solve the equation: $$2\arctan(x)=\arcsin(x)+\arccos(x)=\frac{\pi}{2}$$ $$\arctan(x)=\frac{\pi}{4}$$
I leave the rest to you.