# Prove that : $$sin^{-1}{12\over 13}$$ + $$cot^{-1}{4\over 3}$$ + $$tan^{-1}{63\over 16}$$ = $$\pi$$

## Solution :

We have, $$sin^{-1}{12\over 13}$$ + $$cot^{-1}{4\over 3}$$ + $$tan^{-1}{63\over 16}$$

= $$tan^{-1}{12\over 5}$$ + $$tan^{-1}{3\over 4}$$ + $$tan^{-1}{63\over 16}$$

= $$\pi$$ + $$tan^{-1}({{{12\over 5} + {3\over 4}}\over {1 – {12\over 5} \times {3\over 4}}})$$ + $$tan^{-1}{63\over 16}$$

= $$\pi$$ + $$tan^{-1}{63\over (-16)}$$ + $$tan^{-1}{63\over 16}$$

= $$\pi$$ – $$tan^{-1}{63\over 16}$$ + $$tan^{-1}{63\over 16}$$

= $$\pi$$

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