Question
$(\text{a}-\text{b})\cos\frac{\text{C}}{2}=\text{c}\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)$

Answer

$(\text{a}-\text{b})\cos\frac{\text{C}}{2}=\text{c}\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)$ Let $\text{a = k}\sin\text{A, b = k}\sin\text{B,c = k}\sin\text{C}$ $\text{LHS}=\text{(a}-\text{b})\cos\frac{\text{C}}{2}$ $=\text{k}(\sin\text{A}-\sin\text{B}).\cos\frac{\text{C}}{2}$ $=2\text{k}\cos\Big(\frac{\text{A + B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big).\cos\frac{\text{C}}{2}$ $=2\text{k}\cos\Big(\frac{\pi-\text{C}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big).\cos\frac{\text{C}}2{}$ $=2\text{k}\sin\Big(\frac{\text{C}}{2}\Big).\cos\frac{\text{C}}2{}.\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)$ $\Big[\cos\Big(\frac{\pi}{2}-\theta\Big)=\sin\theta\Big]$ $=\text{k}\sin\text{C}.\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)$ $=\text{c}.\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)=\text{RHS}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free