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Exercise 10(a) · Q5

Q.Show that a+bc=cos⁡A−B2sin⁡C2\dfrac{a+b}{c}=\dfrac{\cos\frac{A-B}{2}}{\sin\frac{C}{2}}.

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Step 1. By the Sine Rule, a=2Rsin⁡A, b=2Rsin⁡B, c=2Rsin⁡Ca=2R\sin A,\ b=2R\sin B,\ c=2R\sin C. So

a+bc=sin⁡A+sin⁡Bsin⁡C.\frac{a+b}{c}=\frac{\sin A+\sin B}{\sin C}.

Step 2. Apply the sum-to-product identity to the numerator:

sin⁡A+sin⁡B=2sin⁡A+B2cos⁡A−B2.\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2}.

Step 3. Since A+B+C=180∘A+B+C=180^\circ, we have A+B2=90∘−C2\frac{A+B}{2}=90^\circ-\frac{C}{2}, so sin⁡A+B2=sin⁡(90∘−C2)=cos⁡C2\sin\frac{A+B}2=\sin\left(90^\circ-\frac C2\right)=\cos\frac C2. Also write sin⁡C=2sin⁡C2cos⁡C2\sin C=2\sin\frac C2\cos\frac C2 (double angle). Then …

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