Question

If $\cot \alpha \cot \beta =2$ , then $ \frac{\cos\left(\alpha +\beta\right)}{\cos\left(\alpha-\beta\right)} $ is equal to

• A. 3
• B. $ \frac{2}{3} $
• C. $ \frac{1}{3} $
• D. $\tan \alpha \, \tan \beta $

Answer 1

Answer: (C)

$ \frac{\cos\left(\alpha +\beta\right)}{\cos\left(\alpha-\beta\right)} = \frac{\cos\alpha \cos \beta -\sin \alpha \sin \beta }{\cos \alpha \cos \beta + \sin \alpha \sin \beta}$
Dividing numerator and denominator by $\sin \alpha \, \sin \beta$
$\frac{\cot \alpha \cot \beta-1}{ \cot \alpha \cot \beta + 1} = \frac{2-1}{2+1}=\frac{1}{3} $