How do I draw the graphs of $\sin^{-1}(\frac{2x}{1+x^2})$ and $\tan^{-1}(\frac{2x}{1-x^2})$? The solution of a question in my books how these graphs and draws the conclusion that from them it is seen $0<x<1$.How do I draw these graphs and how is this conclusion obtained?
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Set $y=x$ in https://math.stackexchange.com/questions/672575/proof-for-the-formula-of-sum-of-arcsine-functions-arcsin-x-arcsin-y and https://math.stackexchange.com/questions/1837410/inverse-trigonometric-function-identity-doubt-tan-1x-tan-1y-pi-tan – lab bhattacharjee May 06 '19 at 08:42
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1Can you clarify the question? $0 < x < 1$ for positive real $x$ is not true for the arctangent function. – Toby Mak May 06 '19 at 08:43
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@TobyMak if a and b satisfy sin^-1 (2x/1+x^2) = tan^-1 (2x/1-x^2) and |a-b| < k then k =? Answer is 2. – Hema May 06 '19 at 10:15
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The second claim is not necessarily true. The first function is defined for all real $x,$ and the second is defined for all $|x|\ne 1.$ – Allawonder May 06 '19 at 11:20