Question

If $\sin A=\sin B$ and $\cos A=\cos B$, then

A.$\sin \frac{A-B}{2}=0$

B.$\sin \frac{A+B}{2}=0$

C.$\cos \frac{A-B}{2}=0$

D.$\cos A+B=0$

• A.
• B.
• C.
• D.

Answer 1

The correct option is A .$\sin \frac{A-B}{2}=0$

We have $\sin A=\sin B$ and $\cos A=\cos B$

Using $\sin 2\theta =2\sin \theta \cos \theta$, we get,

$\Rightarrow 2\sin \frac{A}{2}\cos \frac{A}{2}=2\sin \frac{B}{2}\cos \frac{B}{2}$

$\Rightarrow \sin \frac{A}{2}\cos \frac{A}{2}=\sin \frac{B}{2}\cos \frac{B}{2}$..(i)

Using $\cos 2\theta =2{\cos }^2\theta -1$, we get,

$\Rightarrow 2{\cos }^2\frac{A}{2}-1=2{\cos }^2\frac{B}{2}-1$

$\Rightarrow \cos \frac{A}{2}=\cos \frac{B}{2}$...(ii)

From (i) and (ii)

$\Rightarrow \sin \frac{A}{2}=\sin \frac{B}{2}$ ...(iii)

Consider, $\sin \left(\frac{A-B}{2}\right)$

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

$=0$ [From (ii) and (iii)]

Hence, $\sin \left(\frac{A-B}{2}\right)=0$