Question:
Prove that
$\frac{\cos x+\cos y}{\cos y-\cos x}=\cot \left(\frac{x+y}{2}\right) \cot \left(\frac{x-y}{2}\right)$
Solution:
$=\frac{\cos x-\cos y}{\cos y-\cos x}$
$=\frac{2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}}{-2 \sin \frac{x+y}{2} \sin \frac{y-x}{2}}$
$=\frac{2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}}{2 \sin \frac{x+y}{2} \sin \frac{x-y}{2}}$
$=\frac{\cos \frac{x+y}{2} \cos \frac{x-y}{2}}{\sin \frac{x+y}{2} \sin \frac{x-y}{2}}$
$=\cot \frac{x+y}{2} \cot \frac{x-y}{2}$
Using the formula,
$\cos A-\cos B=-2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}$
$\cos A+\cos B=2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$
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