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Sine And Cosine Formulae And Their Applications — Maths STD 11 Science — Question

Maharashtra 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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