Question

${\left(\frac{\cos A+\cos B}{\sin A-\sin B}\right)}^n+{\left(\frac{\sin A+\sin B}{\cos A-\cos B}\right)}^n=x{\cot }^n\frac{A-B}{2}$ or a, accordingly as n is even or odd. Find $x+a$

Answer 1

L.H.S. $={\left(2\frac{\cos \frac{A+B}{2}\cos \frac{A-B}{2}}{2\cos \frac{A+B}{2}\sin \frac{A-B}{2}}\right)}^n+{\left(\frac{2\sin \frac{A+B}{2}\cos \frac{A-B}{2}}{\sin \frac{A+B}{2}\sin \frac{B-A}{2}}\right)}^n$
$={\left(\cot \frac{A-B}{2}\right)}^n+{(-\cot \frac{A-B}{2})}^n$ $[\because \sin (-\theta )=-\sin \theta ]$
$={\cot }^n\frac{A-B}{2}+{(-1)}^n{\cot }^n\frac{A-B}{2}={\cot }^n\frac{A-B}{2}[1+{(-1)}^n]$
$=0$, if $n$ is odd
$2{\cot }^n\frac{A-B}{2}$, if n is even.
Therefore. $x+a=2+0=2$
Ans: 2