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14)\[\text { Prove that } \frac{\cot A \cot 4 A+1}{\cot A \cot 4 A-1}=\frac{\cos 3 A}{\cos 5 A}\] |
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Answer» \(\frac{cot\,A\,cot\,4A+1}{cot\,A\,cot\,4A-1}\) \(=\frac{cot\,A\,cot\,4A+1}{cot\,A-cot\,4A}\) \(\times\frac{cot\,A+cot\,4A}{cot\,A\,cot\,4A-1}\) \(\times\frac{cot\,A-cot\,4A}{cot\,A+cot\,4A}\) \(=\frac{cot(4A-A)}{cot(4A+A)}\) \(\times\frac{cos\,A\,sin\,4A-cos\,4A\,sin\,A}{cos\,A\,sin\,4A+cos\,4A\,sin\,A}\) \(=\frac{cot\,3A}{cot\,5A}\) \(\times\frac{sin(4A-A)}{sin(4A+A)}\) \(=\frac{cot\,3A}{cot\,5A}\) \(\times\frac{sin\,3A}{sin\,5A}\) \(=\frac{cos\,3A}{sin\,3A}\) \(\times\frac{sin\,5A}{cos\,5A}\)\(\times\frac{sin\,3A}{sin\,5A}\) \(=\frac{cos\,3A}{cos\,5A}\) Hence proved. |
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