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Exercise 3.3 · Q58

Q.Prove the following: 2 cosec 2x+cosec x=sec⁡xcot⁡x22\,\text{cosec}\,2x+\text{cosec}\,x=\sec x\cot\dfrac{x}{2}

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Step 1: 2 cosec 2x+cosec x=22sin⁡xcos⁡x+1sin⁡x=1sin⁡xcos⁡x+1sin⁡x=1+cos⁡xsin⁡xcos⁡x2\,\text{cosec}\,2x+\text{cosec}\,x=\dfrac{2}{2\sin x\cos x}+\dfrac{1}{\sin x}=\dfrac{1}{\sin x\cos x}+\dfrac{1}{\sin x}=\dfrac{1+\cos x}{\sin x\cos x}.

Step 2: Use 1+cos⁡x=2cos⁡2x21+\cos x=2\cos^2\dfrac{x}{2} and sin⁡x=2sin⁡x2cos⁡x2\sin x=2\sin\dfrac{x}{2}\cos\dfrac{x}{2}: 2cos⁡2x2(2sin⁡x2cos⁡x2)cos⁡x=cos⁡x2sin⁡x2cos⁡x\dfrac{2\cos^2\frac{x}{2}}{\left(2\sin\frac{x}{2}\cos\frac{x}{2}\right)\cos x}=\dfrac{\cos\frac{x}{2}}{\sin\frac{x}{2}\cos x}. …

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