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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 Applications4 Marks
Question
$\frac{\text{c}}{\text{a}-\text{b}}=\frac{\tan\big(\frac{\text{A}}{2}\big)+\tan\big(\frac{\text{B}}{2}\big)}{\tan\big(\frac{\text{A}}{2}\big)-\tan\big(\frac{\text{B}}{2}\big)}$
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Answer

$\frac{\text{c}}{\text{a}-\text{b}}=\frac{\tan\big(\frac{\text{A}}{2}\big)+\tan\big(\frac{\text{B}}{2}\big)}{\tan\big(\frac{\text{A}}{2}\big)-\tan\big(\frac{\text{B}}{2}\big)}$$\text{LHS}=\frac{\text{c}}{\text{a}-\text{b}}$
$=\frac{\text{k}\sin\text{C}}{\text{k}\sin\text{A}-\text{k}\sin\text{B}}$
$=\frac{\text{}\sin\text{C}}{\text{}\sin\text{A}-\text{}\sin\text{B}}$
$=\frac{2\sin\frac{\text{C}}{2}\cos\frac{\text{C}}{2}}{\sin\text{A}-\sin\text{B}}$
$=\frac{2\sin\frac{\text{C}}{2}\cos\frac{\text{C}}{2}}{2\cos\big(\frac{\text{A +B}}{2}\big)\sin\big(\frac{\text{A}-\text{B}}{2}\big)}$
$=\frac{\sin\frac{\text{C}}{2}\cos\frac{(\pi-(\text{A + B})}{2}}{\cos\big(\frac{\pi-\text{C}}{2}\big)\sin\big(\frac{\text{A}-\text{B}}{2}\big)}$
$=\frac{\sin\frac{\text{C}}{2}\sin\frac{(\text{A + B})}{2}}{\sin\frac{\text{C}}{2}\sin\big(\frac{\text{A}-\text{B}}{2}\big)}$
$=\frac{\sin\frac{\text{(A + B)}}{2}}{\sin\frac{\text{(A - B)}}{2}}$
$=\frac{\sin\big(\frac{\text{A}}{2}\big).\cos\big(\frac{\text{B}}{2}\big)+\sin\big(\frac{\text{B}}{2}\big).\cos\big(\frac{\text{A}}{2}\big)}{\sin\big(\frac{\text{A}}{2}\big).\cos\big(\frac{\text{B}}{2}\big)-\sin\big(\frac{\text{B}}{2}\big).\cos\big(\frac{\text{A}}{2}\big)}$
$=\frac{\tan\big(\frac{\text{A}}{2}\big)+\tan\big(\frac{\text{B}}{2}\big)}{\tan\big(\frac{\text{A}}{2}\big).-\tan\big(\frac{\text{B}}{2}\big).}=\text{RHS}$ $\Big[$Dividing both Numerator and Denominator by $\cos\big(\frac{\text{A}}{2}\big).\cos\big(\frac{\text{B}}{2}\big)\Big]$

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