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Exercise 3.2 · Q33

Q.Prove the following: cos⁡(π+x)cos⁡(−x)sin⁡(π−x)cos⁡(π2+x)=cot⁡2x\dfrac{\cos(\pi+x)\cos(-x)}{\sin(\pi-x)\cos\left(\dfrac{\pi}{2}+x\right)}=\cot^2x

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Step 1: Reduce each factor: cos⁡(π+x)=−cos⁡x\cos(\pi+x)=-\cos x, cos⁡(−x)=cos⁡x\cos(-x)=\cos x, sin⁡(π−x)=sin⁡x\sin(\pi-x)=\sin x, cos⁡(π2+x)=−sin⁡x\cos\left(\dfrac{\pi}{2}+x\right)=-\sin x.

Step 2: Numerator: cos⁡(π+x)cos⁡(−x)=(−cos⁡x)(cos⁡x)=−cos⁡2x\cos(\pi+x)\cos(-x)=(-\cos x)(\cos x)=-\cos^2x. …

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