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Exercise 3.1 · Q17

Q.Prove that: cot⁡Acot⁡4A+1cot⁡Acot⁡4A−1=cos⁡3Acos⁡5A\dfrac{\cot A\cot4A+1}{\cot A\cot4A-1}=\dfrac{\cos3A}{\cos5A}

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Step 1: Start from the RHS: cos⁡3A=cos⁡(4A−A)=cos⁡4Acos⁡A+sin⁡4Asin⁡A\cos3A=\cos(4A-A)=\cos4A\cos A+\sin4A\sin A and cos⁡5A=cos⁡(4A+A)=cos⁡4Acos⁡A−sin⁡4Asin⁡A\cos5A=\cos(4A+A)=\cos4A\cos A-\sin4A\sin A.

Step 2: Divide both by sin⁡Asin⁡4A\sin A\sin4A: numerator of cos⁡3A\cos3A becomes cot⁡Acot⁡4A+1\cot A\cot4A+1, numerator of cos⁡5A\cos5A becomes cot⁡Acot⁡4A−1\cot A\cot4A-1 (after dividing top and bottom of the ratio by sin⁡Asin⁡4A\sin A\sin4A). …

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