Question

If tan $\alpha =K\cot \beta ,$ then $\frac{\cos (\alpha -\beta )}{\cos (\alpha -\beta )}$ equals

• A. $\frac{1+K}{1-K}$
• B. $\frac{1-K}{1+K}$
• C. $\frac{K+1}{K-1}$
• D. $\frac{K-1}{K+1}$

Answer 1

The correct option is A $\frac{1+K}{1-K}$

Given $\tan \alpha =Kcot\beta$ then, $\frac{\cos (\alpha -\beta )}{\cos (\alpha +\beta )}$

$\Rightarrow \frac{\frac{\sin \alpha }{\cos \alpha }}{\frac{\cos \beta }{\sin \beta }}=K$

$\Rightarrow K=\frac{\sin \alpha \sin \beta }{\cos \alpha \cos \beta }$ ………$\left(1\right)$

Now we have $\frac{\cos (\alpha -\beta )}{\cos (\alpha +\beta )}$ $\left[\begin{array}{c}\cos (a+b)=\cos a\cos b-\sin a\sin b\\ \cos (a-b)=\cos a\cos b+\sin a\sin b\end{array}\right]$

$\Rightarrow \frac{\cos \alpha \cos \beta +\sin \alpha \sin \beta }{\cos \alpha \cos \beta -\sin \alpha \sin \beta }$

On dividing numerator & denominator by $\cos \alpha \cos \beta$ we get

$\Rightarrow \frac{1+\frac{\sin \alpha \sin \beta }{\cos \alpha \cos \beta }}{1-\frac{\sin \alpha \sin \beta }{\cos \alpha \cos \beta }}$

now from equation $\left(1\right)$

$\Rightarrow \frac{1+K}{1-K}$

Hence $[\frac{\cos (\alpha -\beta )}{\cos (\alpha +\beta )}=\frac{1+K}{1-K}]$.