Q. The value of $\left( \frac{\cos{A + \cos B}}{\sin{A - \sin B}} \right)^{n} + \left( \frac{\sin{A + \sin B}}{\cos{A - \cos B}} \right)^{n}$ (where $n\$is an even) is

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

(b)
$\left( \frac{\cos{A + \cos B}}{\sin{A - \sin\ B}} \right)^{n} + \left( \frac{\sin{A + \sin B}}{\cos{A - \cos B}} \right)^{n}$
$\left\lbrack \frac{2\cos{\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right)}}{2\cos{\left( \frac{A + B}{2} \right)\sin\left( \frac{A - B}{2} \right)}} \right\rbrack^{n} + \left\lbrack \frac{2\sin{\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right)}}{2\sin{\left( \frac{A + B}{2} \right)\sin\left( \frac{B - A}{2} \right)}} \right\rbrack^{n}$
${= cot}^{n}{\left( \frac{A - B}{2} \right) + \cot^{n}\left( \frac{B - A}{2} \right)}$
$= \cot^{n}{\left( \frac{A - B}{2} \right) + ( - 1)^{n}\cot\left( \frac{A - B}{2} \right)}$
$= 2\cot^{n}\left( \frac{A - B}{2} \right)\ \ \ \ (\because n$ is even)

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.