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Exercise 2.2 · Q21

Q.If tan⁡θ=12\tan\theta = \dfrac{1}{2}, evaluate 2sin⁡θ+3cos⁡θ4cos⁡θ+3sin⁡θ\dfrac{2\sin\theta + 3\cos\theta}{4\cos\theta + 3\sin\theta}.

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✓ Free question

Step 1. Divide top and bottom by cos⁡θ\cos\theta: 2sin⁡θ+3cos⁡θ4cos⁡θ+3sin⁡θ=2tan⁡θ+34+3tan⁡θ\dfrac{2\sin\theta+3\cos\theta}{4\cos\theta+3\sin\theta} = \dfrac{2\tan\theta+3}{4+3\tan\theta}.

Step 2. Substitute tan⁡θ=12\tan\theta=\tfrac12: 2(1/2)+34+3(1/2)=1+34+1.5=45.5=811\dfrac{2(1/2)+3}{4+3(1/2)} = \dfrac{1+3}{4+1.5} = \dfrac{4}{5.5} = \dfrac8{11}.

✓Final answer

811\dfrac8{11}.

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