c a+b = 1- ( A 2 ) ( B 2 ) 1+ ( A 2 ) ( B 2 )

Question

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

✓

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 $\cos\big(\frac{\text{A}}{2}\big).\cos\big(\frac{\text{B}}{2}\big)\Big]$