Question

The expression ${\left(\frac{\cos A+\cos B}{\sin A-\sin B}\right)}^m+{\left(\frac{\sin A+\sin B}{\cos A-\cos B}\right)}^m$ where $m\in N$, has the value

• A. $2{\cot }^m\left(\frac{A-B}{2}\right)$, if $m$ is odd
• B. $0$, if $m$ is odd
• C. $2{\cot }^m\left(\frac{A-B}{2}\right)$, if $m$ is even
• D. $0$, if $m$ is even

Answer 1

Answer: (B), (C)

The correct options are
B $0$, if $m$ is odd
C $2{\cot }^m\left(\frac{A-B}{2}\right)$, if $m$ is even

Hence

$\frac{\cos A+\cos B}{\sin A-\sin B}$

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

$=\cot \left(\frac{A-B}{2}\right)...\left(i\right)$

Similarly

$\frac{\sin A+\sin B}{\cos A-\cos B}$

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

$=-\cot \left(\frac{A-B}{2}\right).....\left(ii\right)$

Hence if m is odd, the answer is zero.

If $m$ is even, we will get

$2{\left[\cot \left(\frac{A-B}{2}\right)\right]}^m$