{left( {frac{{cos A + cos B}}{{sin A - sin B} | vedclass

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(C) Using the sum-to-product formulas:$\cos A + \cos B = 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$$\sin A - \sin B = 2 \c

Vedclass · Accepted Answer

$0$ if $n$ is odd,$2 \cot^n \left( \frac{A - B}{2} \right)$ if $n$ is even

Vedclass · Answer

(C) Using the sum-to-product formulas:$\cos A + \cos B = 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$$\sin A - \sin B = 2 \cos \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right)$$\sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$$\cos A - \cos B = -2 \sin \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right)$Substituting these into the expression:First term: $\left( \frac{2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}}{2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}} \right)^n = \cot^n \left( \frac{A-B}{2} \right)$Second term: $\left( \frac{2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}}{-2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}} \right)^n = (-\cot \left( \frac{A-B}{2} \right))^n$Total expression = $\cot^n \left( \frac{A-B}{2} \right) + (-1)^n \cot^n \left( \frac{A-B}{2} \right)$If $n$ is even,$(-1)^n = 1$,so the result is $2 \cot^n \left( \frac{A-B}{2} \right)$.If $n$ is odd,$(-1)^n = -1$,so the result is $0$.

${\left( {\frac{{\cos A + \cos B}}{{\sin A - \sin B}}} \right)^n} + {\left( {\frac{{\sin A + \sin B}}{{\cos A - \cos B}}} \right)^n}$ ($n$ is even or odd) $=$

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