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

Question

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

✓

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