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Exercise 10.4 · Q12

Q.Find the derivative of the following: cos⁡ ⁣(2tan⁡−11−x1+x)\cos\!\left(2\tan^{-1}\sqrt{\dfrac{1-x}{1+x}}\right)

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Step 1. Let t=tan⁡−11−x1+xt = \tan^{-1}\sqrt{\dfrac{1-x}{1+x}}, so tan⁡2t=1−x1+x\tan^2 t = \dfrac{1-x}{1+x}.

Step 2. Use the double-angle cosine identity:

cos⁡2t=1−tan⁡2t1+tan⁡2t=1−1−x1+x1+1−x1+x=2x1+x21+x=x\cos 2t = \dfrac{1-\tan^2t}{1+\tan^2t} = \dfrac{1-\frac{1-x}{1+x}}{1+\frac{1-x}{1+x}} = \dfrac{\frac{2x}{1+x}}{\frac{2}{1+x}} = x …

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