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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 Applications4 Marks
Question
$\frac{\text{c}}{\text{a}+\text{b}}=\frac{1-\tan\big(\frac{\text{A}}{2}\big)\tan\big(\frac{\text{B}}{2}\big)}{1+\tan\big(\frac{\text{A}}{2}\big)\tan\big(\frac{\text{B}}{2}\big)}$
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Answer

$\frac{\text{c}}{\text{a}+\text{b}}=\frac{1-\tan\big(\frac{\text{A}}{2}\big)\tan\big(\frac{\text{B}}{2}\big)}{1+\tan\big(\frac{\text{A}}{2}\big)\tan\big(\frac{\text{B}}{2}\big)}$ $\text{LHS}=\frac{\text{c}}{\text{a}+\text{b}}$ $=\frac{\text{k}\sin\text{C}}{\text{k}\sin\text{A}+\text{k}\sin\text{B}}$ $=\frac{2\sin\frac{\text{C}}{2}\cos\frac{\text{C}}{2}}{2\sin\big(\frac{\text{A +B}}{2}\big).\cos\big(\frac{\text{A}-\text{B}}{2}\big)}$ $=\frac{\sin\frac{\text{C}}{2}\cos\frac{\text{C}}{2}}{\sin\big(\frac{\pi-\text{C}}{2}\big).\cos\big(\frac{\text{A}-\text{B}}{2}\big)}$ $=\frac{\sin\big(\frac{\pi-\text{(A + B})}{2}\big)\cos\frac{\text{C}}{2}}{\cos\big(\frac{\text{C}}{2}\big).\cos\big(\frac{\text{A}-\text{B}}{2}\big)}$ $=\frac{\cos\big(\frac{\text{A + B}}{2}\big)}{\cos\big(\frac{\text{A}-\text{B}}{2}\big)}$ $=\frac{\cos\frac{\text{A}}{2}.\cos\frac{\text{B}}{2}-\sin\frac{\text{A}}{2}\sin\frac{\text{B}}{2}}{\cos\frac{\text{A}}{2}.\cos\frac{\text{B}}{2}+\sin\frac{\text{A}}{2}.\sin\frac{\text{B}}2{}}$ $=\frac{1-\tan\frac{\text{A}}{2}\tan\frac{\text{B}}2{}}{1+\tan\frac{\text{A}}{2}.\tan\frac{\text{B}}{2}}=\text{RHS}$ $\Big[$Dividing both Numberator and Denomiator by $\cos\big(\frac{\text{A}}{2}\big).\cos\big(\frac{\text{B}}{2}\big)\Big]$

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