Prove that 1/2tan-1x=cos-1√1+√1+x2/2√1+x2
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Posted by Dishaant Pandit 7 years, 1 month ago
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Naveen Sharma 7 years, 1 month ago
Ans. \(\)
\({1\over 2}tan^{-1}x = cos^{-1} (\sqrt {1 +\sqrt{1+x^2}\over 2\sqrt{1+x^2}})\)
\(taking \space LHS, put \space x = tan \theta \)
=> \( cos^{-1} (\sqrt {1 +\sqrt{1+tan^2\theta}\over 2\sqrt{1+tan^2\theta}})\)
=> \( cos^{-1} (\sqrt {1 +\sqrt{sec^2\theta}\over 2\sqrt{sec^2\theta}})\)
=> \( cos^{-1} (\sqrt {1 +{sec\theta}\over 2{sec\theta}})\)
=> \( cos^{-1} ({\sqrt {1 +cos\theta\over 2}})\)
=> \( cos^{-1} ({\sqrt { cos^2{\theta\over 2}}})\)
=> \( cos^{-1} ({ cos{\theta\over 2}})\)
=> \(LHS = {\theta\over 2}\)
\(Taking \space RHS, put \space x = tan\theta \)
=> \({1\over 2} tan^{-1}({tan \theta })\)
=> \(RHS = {{\theta \over 2}}\)
RHS = LHS
Hence Proved
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