If `tanalpha=kcotbeta`, then `(cos(alpha-beta))/(cos(alpha+beta))` is equal to

Options

A. `(1+k)/(1-k)`
B. `(1-k)/(1+k)`
C. `(k+1)/(k-1)`
D. `(k-1)/(k+1)`

Explanation

To solve the problem, we start with the given equation:

\[ \tan \alpha = k \cot \beta \]

We can rewrite \(\cot \beta\) in terms of \(\tan \beta\):

\[ \tan \alpha = k \cdot \frac{1}{\tan \beta} \implies \tan \alpha \tan \beta = k \]

Now we need to find the expression:

\[ \frac{\cos(\alpha - \beta)}{\cos(\alpha + \beta)} \]

Using the cosine of sum and difference formulas, we have:

\[ \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \]
\[ \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \]

Substituting these into our expression gives:

\[ \frac{\cos(\alpha - \beta)}{\cos(\alpha + \beta)} = \frac{\cos \alpha \cos \beta + \sin \alpha \sin \beta}{\cos \alpha \cos \beta - \sin \alpha \sin \beta} \]

Next, we can divide the numerator and the denominator by \(\cos \alpha \cos \beta\):

\[ = \frac{1 + \frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta}}{1 - \frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta}} = \frac{1 + \tan \alpha \tan \beta}{1 - \tan \alpha \tan \beta} \]

Since we know that \(\tan \alpha \tan \beta = k\), we can substitute \(k\) into the expression:

\[ = \frac{1 + k}{1 - k} \]

Thus, the final result is:

\[ \frac{\cos(\alpha - \beta)}{\cos(\alpha + \beta)} = \frac{1 + k}{1 - k} \]