Solution : Here, 2\(tan^{-1}({2x+1})\) = \(cos^{-1}x\) cos(2\(tan^{-1}({2x+1})\)) = x  { We Know cos2x = \({1-tan^2x\over {1+tan^2x}}\)} \(\therefore\)  \({{1-{(2x+1)}^2}\over {1-{(2x+1)}^2}}\) = x   \(\implies\)   (1 – 2x – 1)(1 + 2x + 1) = x(\(4x^2 + 4x + 2\)) \(\implies\)  -2x.2(x + 1) = 2x(\(2x^2 + 2x + 1\))  \(\implies\)  2x(\(2x^2 + 2x + 1 + 2x

Solution : We have, \(sin^{-1}{12\over 13}\) + \(cot^{-1}{4\over 3}\) + \(tan^{-1}{63\over 16}\) = \(tan^{-1}{12\over 5}\) + \(tan^{-1}{3\over 4}\) + \(tan^{-1}{63\over 16}\) = \(\pi\) + \(tan^{-1}({{{12\over 5} + {3\over 4}}\over {1 – {12\over 5} \times {3\over 4}}})\) + \(tan^{-1}{63\over 16}\) = \(\pi\) + \(tan^{-1}{63\over (-16)}\) + \(tan^{-1}{63\over 16}\) = \(\pi\) – \(tan^{-1}{63\over 16}\) + \(tan^{-1}{63\over 16}\)

Solution : We know that \(sin^{-1}(sinx)\) = x, if \(-\pi\over 2\) \(\le\) x \(\le\) \(\pi\over 2\) Here, x = 10 radians which does not lie between -\(\pi\over 2\) and \(\pi\over 2\) But, \(3\pi\) – x i.e. \(3\pi\) – 10 lie between -\(\pi\over 2\) and \(\pi\over 2\) Also, sin(\(3\pi\) – 10) = sin 10 \(\therefore\)  \(sin^{-1}(sin10)\)

Solution : We have, L.H.S. = \(cos^{-1}{12\over 13}\) + \(sin^{-1}{3\over 5}\) = \(tan^{-1}{5\over 12}\) + \(tan^{-1}{3\over 4}\) \(\because\) [ \(cos^{-1}{12\over 13}\) = \(tan^{-1}{5\over 12}\) & \(sin^{-1}{3\over 5}\) = \(tan^{-1}{3\over 4}\) ] L.H.S. = \(tan^{-1}({{{5\over 12} + {3\over 4}}\over {1 – {5\over 12}.{3\over 4}}})\) = \(tan^{-1}{56\over 33}\) R.H.S. = \(sin^{-1}{56\over 65}\) = \(tan^{-1}{56\over 33}\) L.H.S =

Solution : \(sin^{-1}({-\sqrt{3}\over 2})\) = – \(sin^{-1}({\sqrt{3}\over 2})\) = \(-\pi\over 3\) \(cos^{-1}(cos({7\pi\over 6}))\) = \(cos^{-1}(cos({2\pi – {5\pi\over 6}}))\) = \(cos^{-1}(cos({5\pi\over 6}))\) = \(5\pi\over 6\) Hence \(sin^{-1}({-\sqrt{3}\over 2})\) + \(cos^{-1}(cos({7\pi\over 6}))\) = \(-\pi\over 3\) + \(5\pi\over 6\) = \(\pi\over 2\) Similar Questions Solve the equation : 2\(tan^{-1}({2x+1})\) = \(cos^{-1}x\) Prove that : \(sin^{-1}{12\over 13}\) + \(cot^{-1}{4\over