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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