Question

lf n is an odd integer, then ${\left(\frac{\cos A+\cos B}{\sin A-\sin B}\right)}^n+{\left(\frac{\sin A+\sin B}{\cos A-\cos B}\right)}^n$ is equal to

• A. $0$
• B. $2{\cot }^n\left(\frac{A-B}{2}\right)$
• C. $2{\tan }^n\left(\frac{A+B}{2}\right)$
• D. None of these

Answer 1

The correct option is A $0$

${\left(\frac{\cos A+\cos B}{\sin A-\sin B}\right)}^n+{\left(\frac{\sin A+\sin B}{\cos A-\cos B}\right)}^n$

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

$={\left(\cot \left(\frac{A-B}{2}\right)\right)}^n+{(-\cot \left(\frac{A-B}{2}\right))}^n$

$=0$
as n is odd