<s>[INST] Prove that $\cos^{-1}(\frac{2}{\sqrt 7})+\tan^{-1}(\frac{\sqrt 3}{5}) = \frac{\pi}{3}$ I started off by making two right angled triangles. One having an angle $\alpha$ such that $\cos \alpha = \frac{2}{\sqrt 7}$ and the other having an angle $\beta$ such that $\tan \beta = \frac{\sqrt 3}{5}$.
Now from here I must show that $\alpha + \beta = \frac{\pi}{3}$.
However I am unable to find a relation between these two triangles.
It would be appreciated if someone could help me with the proof or give some kind of hint.
Thank you in advance.
 [/INST] Observe that$$\cos^{-1}\left(\frac{2}{\sqrt 7} \right) = \tan^{-1}\left(\frac{\sqrt3}{2} \right)$$
So we have
$$\tan^{-1}\left(\frac{\sqrt3}{2} \right) + \tan^{-1}\left(\frac{\sqrt3}{5} \right) = \tan^{-1}\left(\frac{\frac{\sqrt3}{2} + \frac{\sqrt3}{5}}{1 - \frac{\sqrt3}{2} \cdot \frac{\sqrt3}{5}} \right) = \tan^{-1}(\sqrt 3) = \frac{\pi}{3}$$
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