Question

In $△ABC$,
$\cos \left(\frac{A-B}{2}\right)=\left(\frac{a+b}{c}\right)\sin \frac{C}{2}$

• A. True
• B. False

Answer 1

The correct option is A True

Using sine formula ,

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k\left(say\right)$

$a=k\sin A,b=k\sin B,c=k\sin C$

Now,

$\frac{(a+b)}{c}=\frac{(k\sin A+k\sin B)}{k\sin C}$

$=\frac{2(.\sin \frac{(A+B)}{2}).\left(\cos \frac{(A-B)}{2}\right)}{(2.\sin \frac{C}{2}.\cos \frac{C}{2})}$

$=\frac{\left(\sin \right(\frac{\pi }{2}-\frac{C}{2}).\cos (\frac{A-B}{2}\left)\right)}{(\sin \frac{C}{2}.\cos \frac{C}{2})}$

$=\frac{\cos \frac{C}{2}.\cos \frac{A-B}{2}}{\sin \frac{C}{2}.\cos \frac{C}{2}}$

$=\frac{\cos \frac{A-B}{2}}{sin\frac{C}{2}}$

Hence,

$\frac{a+b}{c}=\frac{\cos \frac{A-B}{2}}{sin\frac{C}{2}}$

So,

$\cos \left(\frac{A-B}{2}\right)=\left(\frac{a+b}{c}\right)\sin \frac{C}{2}$