Question

If $n$ is an odd integer then ${\left(\frac{\cos A+\cos B}{\sin A-\sin B}\right)}^n+{\left(\frac{\sin A+\sin B}{\cos A-\cos B}\right)}^n=$

• A. $0$
• B. ${\cot }^n\left(\frac{A+B}{2}\right)$
• C. ${\cot }^n\left(\frac{A-B}{2}\right)$
• D. $2{\tan }^n\left(\frac{A+B}{2}\right)$

Answer 1

Answer: (A)

$0$

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

If $n$ is even then ${(-1)}^n=1$

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

If $n$ is even then ${(-1)}^n=-1$

$\therefore$ Given expression become $0$