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Exercise 3.3 · Q60

Q.Prove the following: sin⁡xtan⁡x2+2cos⁡x=21+tan⁡2x2\sin x\tan\dfrac{x}{2}+2\cos x=\dfrac{2}{1+\tan^2\dfrac{x}{2}}

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Step 1: Let t=tan⁡x2t=\tan\dfrac{x}{2}, so sin⁡x=2t1+t2\sin x=\dfrac{2t}{1+t^2}, cos⁡x=1−t21+t2\cos x=\dfrac{1-t^2}{1+t^2}.

Step 2: sin⁡xtan⁡x2=2t1+t2⋅t=2t21+t2\sin x\tan\dfrac{x}{2}=\dfrac{2t}{1+t^2}\cdot t=\dfrac{2t^2}{1+t^2}.

Step 3: 2cos⁡x=2(1−t2)1+t22\cos x=\dfrac{2(1-t^2)}{1+t^2}. …

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