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Miscellaneous Exercise 3 (II) · Q116

Q.Prove the following: tan⁡6°tan⁡42°tan⁡66°tan⁡78°=1\tan6°\tan42°\tan66°\tan78°=1

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Step 1: Use tan⁡Atan⁡B=cos⁡(A−B)−cos⁡(A+B)cos⁡(A−B)+cos⁡(A+B)\tan A\tan B=\dfrac{\cos(A-B)-\cos(A+B)}{\cos(A-B)+\cos(A+B)}.

Step 2: For A=66°,B=6°A=66°,B=6°: A−B=60°,A+B=72°A-B=60°,A+B=72°, so tan⁡66°tan⁡6°=cos⁡60°−cos⁡72°cos⁡60°+cos⁡72°=12−cos⁡72°12+cos⁡72°\tan66°\tan6°=\dfrac{\cos60°-\cos72°}{\cos60°+\cos72°}=\dfrac{\frac12-\cos72°}{\frac12+\cos72°}.

Step 3: For A=78°,B=42°A=78°,B=42°: A−B=36°,A+B=120°A-B=36°,A+B=120°, so tan⁡78°tan⁡42°=cos⁡36°−cos⁡120°cos⁡36°+cos⁡120°=cos⁡36°+12cos⁡36°−12\tan78°\tan42°=\dfrac{\cos36°-\cos120°}{\cos36°+\cos120°}=\dfrac{\cos36°+\frac12}{\cos36°-\frac12}. …

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