Aspire Faculty ID #16324 · Topic: NIMCET 2010 · Just now
NIMCET 2010

If $\sin^{-1}\frac{2a}{1+a^2} - \cos^{-1}\frac{1-b^2}{1+b^2} = \tan^{-1}\frac{2x}{1-x^2}$ then $x$ equals:

Solution

$\sin^{-1}\frac{2a}{1+a^2} = 2\tan^{-1} a$ $\cos^{-1}\frac{1-b^2}{1+b^2} = 2\tan^{-1} b$ So equation becomes: $2\tan^{-1} a - 2\tan^{-1} b = \tan^{-1}\frac{2x}{1-x^2}$ Use identity: $\tan^{-1}u - \tan^{-1}v = \tan^{-1}\frac{u-v}{1+uv}$ Thus $2\tan^{-1}\frac{a-b}{1+ab} = \tan^{-1}\frac{2x}{1-x^2}$ This gives: $x=\frac{a-b}{1+ab}$

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