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

Question

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

Dr. Abubacker · Accepted Answer

(a)
Since, $	an\alpha = k\cot\beta$ or $	an{\alpha	an{\beta = k}}$
Now, $\frac{\cos{(\alpha - \beta)}}{\cos{(\alpha + \beta)}} = \frac{\cos\alpha\cos\beta + \sin\alpha\sin\beta}{\cos\alpha\cos\beta - \sin\alpha\sin\beta}$
$= \frac{1 + 	an\alpha	an\beta}{1 - 	an\alpha	an\beta} = \frac{1 + k}{1 - k}$

Q. If $\tan{\alpha = k\cot\beta}$, then $\frac{\cos{(\alpha - \beta)}}{\cos{(\alpha + \beta)}}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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