Q.Prove that $2\tan^{-1}\!\left(\dfrac13\right)=\tan^{-1}\!\left(\dfrac34\right)$.

Using the double-angle form $2\tan^{-1}x=\tan^{-1}\!\left(\dfrac{2x}{1-x^2}\right)$ (valid for $-1<x<1$) with $x=\dfrac13$:

$2\tan^{-1}\!\left(\frac13\right)=\tan^{-1}\!\left(\frac{2\cdot\frac13}{1-\frac19}\right)=\tan^{-1}\!\left(\frac{\frac23}{\frac89}\right)=\tan^{-1}\!\left(\frac23\cdot\frac98\right)=\tan^{-1}\!\left(\frac34\right).$ …

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