# Prove that $\frac{1}{2}\cot^{-1}\frac{2\sqrt[3]{4}+1}{\sqrt{3}}+\frac{1}{3}\tan^{-1}\frac{\sqrt[3]{4}+1}{\sqrt{3}}=\dfrac{\pi}{6}.$

Prove that $$\frac{1}{2}\cot^{-1}\frac{2\sqrt[3]{4}+1}{\sqrt{3}}+\frac{1}{3}\tan^{-1}\frac{\sqrt[3]{4}+1}{\sqrt{3}}=\dfrac{\pi}{6}.$$

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Use the following identities:
$$\arctan a \pm\arctan b=\arctan\left(\frac{a\pm b}{1\mp ab}\right)$$ $$\arctan a+\mathrm{arccot} \hspace{2pt}a=\frac{\pi}{2},\hspace{10 pt}\forall a>0$$

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Let $y^3=4$ and $x^2=3$

So, we need to prove $\frac{1}{2}\cot^{-1}\frac{2y+1}{x}+\frac{1}{3}\tan^{-1}\frac{y+1}{x}=\dfrac{\pi}{6}.$

or $3\cot^{-1}\frac{2y+1}{x}+2\tan^{-1}\frac{y+1}{x}=\pi$

or $3(\frac{\pi}{2}-\tan^{-1}\frac{2y+1}{x})+2\tan^{-1}\frac{y+1}{x}=\pi$ as $\tan^{-1}a+\cot^{-1}a=\frac{\pi}{2}$

or we need to prove, $3\tan^{-1}\frac{2y+1}{x}-2\tan^{-1}\frac{y+1}{x}=\frac{\pi}{2}$

As $2\tan^{-1}a=\tan^{-1}\frac{2a}{1-a^2}$ and $3\tan^{-1}a=\tan^{-1}\frac{3a-a^3}{1-3a^2}$

$2\tan^{-1}\frac{y+1}{x}=\tan^{-1}\frac{2x(y+1)}{x^2-(y+1)^2}$ $=\tan^{-1}\frac{-2x(y+1)}{y^2+2y-2}$ as $x^2=3$

$3\tan^{-1}\frac{2y+1}{x}=\tan^{-1}\frac{3x^2(2y+1)-(2y+1)^3}{x(x^2-3(2y+1)^2)}$ $=\tan^{-1}\frac{y^2-y+2}{x(y^2+y)}$ as $x^2=3$ and $y^3=4$

or $\tan^{-1}\frac{y^2-y+2}{x(y^2+y)}-\tan^{-1}\frac{-2x(y+1)}{y^2+2y-2}$ $=\frac{\pi}{2}$

Now, if $\tan^{-1}a-\tan^{-1}b=\frac{\pi}{2}$, $\tan^{-1}a=\frac{\pi}{2}+\tan^{-1}b=\cot^{-1}(-b)=\tan^{-1}(-\frac{1}{b})$

$\implies ab=-1$

So, we need to show $\frac{y^2-y+2}{x(y^2+y)}\frac{-2x(y+1)}{y^2+2y-2}=-1$ using $x^2=3$ and $y^3=4$

or $\frac{y^2-y+2}{(y^2+y)}\frac{-2(y+1)}{y^2+2y-2}=-1$

As $y^3=4$, $(y+1)(y^2-y+2)=(y+1)(y^2-y+1)+y+1=y^3+1+y+1=6+y$

and $(y^2+y)(y^2+2y-2)$ $=y^4+y^3(2+1)+y^2(2-2)-2y=y^3(y+3)-2y=4(y+3)-2y=2y+12=2(y+6)$

So, $\frac{y^2-y+2}{(y^2+y)}\frac{-2(y+1)}{y^2+2y-2}=\frac{(6+y)(-2)}{2(6+y)}=-1$

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