If `tan theta_(1).tantheta_(2)=k`, then: `(cos(theta_(1)-theta_(2))/(cos(theta_(1)+theta_(2)))=`

Options

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

Answer

To solve the problem, we need to find the expression for \(\frac{\cos(\theta_1 - \theta_2)}{\cos(\theta_1 + \theta_2)}\) given that \(\tan \theta_1 \cdot \tan \theta_2 = k\).

### Step-by-Step Solution:

1. **Recall the Formulas**:
We start with the formulas for \(\cos(\theta_1 - \theta_2)\) and \(\cos(\theta_1 + \theta_2)\):
\[ \cos(\theta_1 - \theta_2) = \cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2 \]
\[ \cos(\theta_1 + \theta_2) = \cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2 \]

2. **Set Up the Fraction**:
We can now set up the fraction:
\[ \frac{\cos(\theta_1 - \theta_2)}{\cos(\theta_1 + \theta_2)} = \frac{\cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2}{\cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2} \]

3. **Divide Numerator and Denominator**:
To simplify, we divide both the numerator and the denominator by \(\cos \theta_1 \cos \theta_2\):
\[ = \frac{\frac{\cos \theta_1 \cos \theta_2}{\cos \theta_1 \cos \theta_2} + \frac{\sin \theta_1 \sin \theta_2}{\cos \theta_1 \cos \theta_2}}{\frac{\cos \theta_1 \cos \theta_2}{\cos \theta_1 \cos \theta_2} - \frac{\sin \theta_1 \sin \theta_2}{\cos \theta_1 \cos \theta_2}} \]
This simplifies to:
\[ = \frac{1 + \frac{\sin \theta_1}{\cos \theta_1} \cdot \frac{\sin \theta_2}{\cos \theta_2}}{1 - \frac{\sin \theta_1}{\cos \theta_1} \cdot \frac{\sin \theta_2}{\cos \theta_2}} \]

4. **Substitute Tangent Values**:
Recognizing that \(\frac{\sin \theta}{\cos \theta} = \tan \theta\), we can substitute:
\[ = \frac{1 + \tan \theta_1 \tan \theta_2}{1 - \tan \theta_1 \tan \theta_2} \]
Given that \(\tan \theta_1 \tan \theta_2 = k\), we substitute \(k\) into the equation:
\[ = \frac{1 + k}{1 - k} \]

5. **Final Result**:
Thus, the expression simplifies to:
\[ \frac{\cos(\theta_1 - \theta_2)}{\cos(\theta_1 + \theta_2)} = \frac{1 + k}{1 - k} \]

### Conclusion:
The final answer is:
\[ \frac{\cos(\theta_1 - \theta_2)}{\cos(\theta_1 + \theta_2)} = \frac{1 + k}{1 - k} \]