Question

$\frac{\sin A-\sin B}{\cos A+\cos B}$ is equal to

• A. $\sin \left(\frac{A+B}{2}\right)$
• B. $2\tan (A+B)$
• C. $\cot \left(\frac{A-B}{2}\right)$
• D. $\tan \left(\frac{A-B}{2}\right)$
• E. $2\cot (A+B)$

Answer 1

Answer: (D)

Given that,
$\frac{\sin A-\sin B}{\cos A+\cos B}$
$=\frac{2\sin \left(\frac{A-B}{2}\right)\cos \left(\frac{A+B}{2}\right)}{2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)}$
$=\frac{\sin \left(\frac{A-B}{2}\right)}{\cos \left(\frac{A-B}{2}\right)}=\tan \left(\frac{A-B}{2}\right)$