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Sine and Cosine Formulae and Their Applications — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSSine and Cosine Formulae and Their Applications2 Marks
Question
In triangle ABC, prove the following: $\frac{\text{a}-\text{b}}{\text{a + b}}=\frac{\tan\big(\frac{\text{A}-\text{B}}{2}\big)}{\tan\big(\frac{\text{A}+\text{B}}{2}\big)}$
✓

Answer

$\frac{\text{a}-\text{b}}{\text{a + b}}=\frac{\tan\big(\frac{\text{A}-\text{B}}{2}\big)}{\tan\big(\frac{\text{A}+\text{B}}{2}\big)}$ Let $\text{a = k}\sin\text{A,b = k}\sin\text{B}$ (Using sine rule) $\text{LHS}=\frac{\text{a}-\text{b}}{\text{a + b}}$ $=\frac{\text{k}\sin\text{A}-\text{k}\sin\text{B}}{\text{k}\sin\text{A}+\text{k}\sin\text{B}}$ $=\frac{\sin\text{A}-\sin\text{B}}{\sin\text{A}+\sin\text{B}}$ $=\frac{2\cos\big(\frac{\text{A +B}}{2}\big)\sin\big(\frac{\text{A}-\text{B}}{2}\big)}{2\sin\big(\frac{\text{A +B}}{2}\big)\cos\big(\frac{\text{A}-\text{B}}{2}\big)}$ $=\frac{\tan\big(\frac{\text{A}-\text{B}}{2}\big)}{\tan\big(\frac{\text{A}+\text{B}}{2}\big)}=\text{RHS}$

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