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Miscellaneous Exercise 3 · Q125

Q.Solve: tan⁡−11−x1+x=12tan⁡−1x\tan^{-1}\dfrac{1-x}{1+x} = \dfrac{1}{2}\tan^{-1}x, for x>0x > 0.

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Using the standard identity tan⁡−11−x1+x=π4−tan⁡−1x\tan^{-1}\dfrac{1-x}{1+x}=\dfrac{\pi}{4}-\tan^{-1}x (for x>−1x>-1, from the tan-subtraction formula with tan⁡(π/4)=1\tan(\pi/4)=1): the equation becomes $\dfrac{\pi}{4}-\tan^{-1}x=\dfrac12\tan^{-1}x\implies\dfrac{\pi}{4}=\dfrac32\tan^{-1}x\im …

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