Inverse trigonometic functions Question:
Prove that $\tan^{-1}(\frac{1}{2} \tan2A)+\tan^{-1}(\cot A)+\tan^{-1}(\cot^{3} A)=0$
MyProblem
We can just use the formula of $\tan^{-1} A +\tan^{-1} B$ but I think it would be a waste of time. Is there any other shorter and simpler method to solve it.
 A: The hint:
Prove that
$$\left(\tan^{-1}(\frac{1}{2} \tan2A)+\tan^{-1}(\cot A)+\tan^{-1}(\cot^{3} A)\right)'=0$$ and check $A=\frac{\pi}{2}.$
Indeed, $$\left(\arctan\left(\frac{1}{2}\tan2x\right)+\arctan\cot{x}+\arctan\cot^3x\right)'=$$
$$=\frac{\frac{1}{\cos^22x}}{1+\frac{1}{4}\tan^22x}+\frac{-\frac{1}{\sin^2x}}{1+\cot^2x}+\frac{3\cot^2x\cdot\left(-\frac{1}{\sin^2x}\right)}{1+\cot^6x}=$$
$$=\frac{4}{4\cos^22x+\sin^22x}-1-\frac{3\sin^2x\cos^2x}{\sin^6x+\cos^6x}=$$
$$=\frac{4}{4-3\sin^22x}-1-\frac{3\sin^2x\cos^2x}{1-3\sin^2x\cos^2x}=$$
$$=\frac{1}{1-3\sin^2x\cos^2x}-1-\frac{3\sin^2x\cos^2x}{1-3\sin^2x\cos^2x}=0.$$
Thus, our expression is a constant on all interval of the domain:
 $$(0,\pi)\setminus\left\{\frac{\pi}{4},\frac{3\pi}{4}\right\},$$
for which it's enough to check $$x\in\left\{\frac{\pi}{8},\frac{\pi}{2},\frac{7\pi}{8}\right\},$$  which gives $\pi$ on $\left(0,\frac{\pi}{4}\right)$, $0$ on $\left(\frac{\pi}{4},\frac{3\pi}{4}\right)$ and $-\pi$ on $\left(\frac{3\pi}{4},\pi\right).$
