Question

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

Answer 1

Hint: We first use the associative angle formulas for the trigonometric ratios. We divide with similar terms. We break the value of $n$ for odd and even terms. We find the value of $x,a$. Then we find the value of $x+a$.

Complete step by step answer:
We are going to use the associative angle formulas for the trigonometric ratios. We have
\[\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\]
$\cos A-\cos B=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{B-A}{2} \right)$
$\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$
$\sin A-\sin B=2\cos \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$
We put these formulas in the left-side of the equation ${{\left( \dfrac{\cos A+\cos B}{\sin A-\sin B} \right)}^{n}}+{{\left( \dfrac{\sin A+\sin B}{\cos A-\cos B} \right)}^{n}}$.
$\begin{align}
  & {{\left( \dfrac{\cos A+\cos B}{\sin A-\sin B} \right)}^{n}}+{{\left( \dfrac{\sin A+\sin B}{\cos A-\cos B} \right)}^{n}} \\
 & ={{\left( \dfrac{2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)}{2\cos \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)} \right)}^{n}}+{{\left( \dfrac{2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)}{2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{B-A}{2} \right)} \right)}^{n}} \\
\end{align}$
We now divide both numerator and denominator with the similar terms like $2\cos \left( \dfrac{A+B}{2} \right)$ and $2\sin \left( \dfrac{A+B}{2} \right)$.
$\begin{align}
  & {{\left( \dfrac{2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)}{2\cos \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)} \right)}^{n}}+{{\left( \dfrac{2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)}{2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{B-A}{2} \right)} \right)}^{n}} \\
 & ={{\cot }^{n}}\left( \dfrac{A-B}{2} \right)+{{\left\{ \dfrac{\cos \left( \dfrac{A-B}{2} \right)}{\sin \left( \dfrac{B-A}{2} \right)} \right\}}^{n}} \\
\end{align}$
We try to keep the angle in the form of $\dfrac{A-B}{2}$ and we also know that $\sin \left( -x \right)=-\sin x$.
So, we have \[\sin \left( \dfrac{B-A}{2} \right)=\sin \left( -\dfrac{A-B}{2} \right)=-\sin \left( \dfrac{A-B}{2} \right)\].
We get \[{{\cot }^{n}}\left( \dfrac{A-B}{2} \right)+{{\left\{ -\dfrac{\cos \left( \dfrac{A-B}{2} \right)}{\sin \left( \dfrac{A-B}{2} \right)} \right\}}^{n}}={{\cot }^{n}}\left( \dfrac{A-B}{2} \right)+{{\left\{ -\cot \left( \dfrac{A-B}{2} \right) \right\}}^{n}}\].
We break the value of the $n$ in two parts to simplify the simplification.
First, we take the value of $n$ to be even. Then we get \[{{\left\{ -\cot \left( \dfrac{A-B}{2} \right) \right\}}^{n}}={{\cot }^{n}}\left( \dfrac{A-B}{2} \right)\].
Now we take the value of $n$ to be odd. Then we get \[{{\left\{ -\cot \left( \dfrac{A-B}{2} \right) \right\}}^{n}}=-{{\cot }^{n}}\left( \dfrac{A-B}{2} \right)\].
Simplifying we get
for $n$ being even, we get ${{\left( \dfrac{\cos A+\cos B}{\sin A-\sin B} \right)}^{n}}+{{\left( \dfrac{\sin A+\sin B}{\cos A-\cos B} \right)}^{n}}=2{{\cot }^{n}}\left( \dfrac{A-B}{2} \right)$.
for $n$ being odd, we get ${{\left( \dfrac{\cos A+\cos B}{\sin A-\sin B} \right)}^{n}}+{{\left( \dfrac{\sin A+\sin B}{\cos A-\cos B} \right)}^{n}}=0$.
It is given ${{\left( \dfrac{\cos A+\cos B}{\sin A-\sin B} \right)}^{n}}+{{\left( \dfrac{\sin A+\sin B}{\cos A-\cos B} \right)}^{n}}=x{{\cot }^{n}}\dfrac{A-B}{2}$ or $a$ accordingly as n is even or odd.

This gives $x=2,a=0$. Therefore, $x+a=2+0=2$.

Note: We need to remember that we can also divide with $2\cos \left( \dfrac{A+B}{2} \right)$ and $2\sin \left( \dfrac{A+B}{2} \right)$.
 assuming that $\cos \left( \dfrac{A+B}{2} \right),\sin \left( \dfrac{A+B}{2} \right)\ne 0$. But if mentioned to find the general solution then the domain changes to $-\infty \le x\le \infty $. In that case we have to use the formula $\dfrac{A+B}{2}\ne n\pi ,n\pi \pm \dfrac{\pi }{2}$.