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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