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Exercise 3.1 · Q22

Q.If tan⁡A=56\tan A=\dfrac{5}{6}, tan⁡B=111\tan B=\dfrac{1}{11}, prove that A+B=π4A+B=\dfrac{\pi}{4}

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Step 1: Apply tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B} with tan⁡A=56,tan⁡B=111\tan A=\dfrac{5}{6},\tan B=\dfrac{1}{11}.

Step 2: Numerator: 56+111=55+666=6166\dfrac{5}{6}+\dfrac{1}{11}=\dfrac{55+6}{66}=\dfrac{61}{66}. Denominator: 1−56⋅111=1−566=61661-\dfrac{5}{6}\cdot\dfrac{1}{11}=1-\dfrac{5}{66}=\dfrac{61}{66}.

Step 3: So tan⁡(A+B)=61/6661/66=1\tan(A+B)=\dfrac{61/66}{61/66}=1. …

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