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Exercise 3.3 · Q53

Q.Prove the following: 1cot⁡A−cot⁡3A−1tan⁡A−tan⁡3A=cot⁡2A\dfrac{1}{\cot A-\cot3A}-\dfrac{1}{\tan A-\tan3A}=\cot2A

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Step 1: tan⁡A−tan⁡3A=sin⁡Acos⁡3A−cos⁡Asin⁡3Acos⁡Acos⁡3A=sin⁡(A−3A)cos⁡Acos⁡3A=−sin⁡2Acos⁡Acos⁡3A\tan A-\tan3A=\dfrac{\sin A\cos3A-\cos A\sin3A}{\cos A\cos3A}=\dfrac{\sin(A-3A)}{\cos A\cos3A}=\dfrac{-\sin2A}{\cos A\cos3A}, so 1tan⁡A−tan⁡3A=−cos⁡Acos⁡3Asin⁡2A\dfrac{1}{\tan A-\tan3A}=\dfrac{-\cos A\cos3A}{\sin2A}.

Step 2: cot⁡A−cot⁡3A=cos⁡Asin⁡3A−sin⁡Acos⁡3Asin⁡Asin⁡3A=sin⁡(3A−A)sin⁡Asin⁡3A=sin⁡2Asin⁡Asin⁡3A\cot A-\cot3A=\dfrac{\cos A\sin3A-\sin A\cos3A}{\sin A\sin3A}=\dfrac{\sin(3A-A)}{\sin A\sin3A}=\dfrac{\sin2A}{\sin A\sin3A}, so 1cot⁡A−cot⁡3A=sin⁡Asin⁡3Asin⁡2A\dfrac{1}{\cot A-\cot3A}=\dfrac{\sin A\sin3A}{\sin2A}. …

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