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Exercise 3.1 · Q15

Q.Prove that: cos⁡27°+sin⁡27°cos⁡27°−sin⁡27°=tan⁡72°\dfrac{\cos27°+\sin27°}{\cos27°-\sin27°}=\tan72°

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Step 1: Divide numerator and denominator of cos⁡27°+sin⁡27°cos⁡27°−sin⁡27°\dfrac{\cos27°+\sin27°}{\cos27°-\sin27°} by cos⁡27°\cos27°: 1+tan⁡27°1−tan⁡27°\dfrac{1+\tan27°}{1-\tan27°}.

Step 2: This matches tan⁡(45°+27°)=tan⁡45°+tan⁡27°1−tan⁡45°tan⁡27°=1+tan⁡27°1−tan⁡27°\tan(45°+27°)=\dfrac{\tan45°+\tan27°}{1-\tan45°\tan27°}=\dfrac{1+\tan27°}{1-\tan27°} since tan⁡45°=1\tan45°=1. …

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