If `(cos(theta_(1)-theta_(2)))/(cos(theta_(1)+theta_(2)))+(cos(theta_(3)+theta_(4)))/(cos(theta_(3)-theta_(4)))=0,` then `tantheta_(1)tan theta_(2)tan theta_(3)tan theta_(4)=`

Options

A. 1
B. 2
C. `-1`
D. none of these

Answer

To solve the equation

\[ \frac{\cos(\theta_1 - \theta_2)}{\cos(\theta_1 + \theta_2)} + \frac{\cos(\theta_3 + \theta_4)}{\cos(\theta_3 - \theta_4)} = 0, \]

we will start by applying the cosine addition and subtraction formulas.

### Step 1: Apply Cosine Formulas

Using the cosine addition and subtraction formulas, we have:

\[ \cos(a - b) = \cos a \cos b + \sin a \sin b, \]
\[ \cos(a + b) = \cos a \cos b - \sin a \sin b. \]

Applying these to our equation:

\[ \frac{\cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2}{\cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2} + \frac{\cos \theta_3 \cos \theta_4 - \sin \theta_3 \sin \theta_4}{\cos \theta_3 \cos \theta_4 + \sin \theta_3 \sin \theta_4} = 0. \]

### Step 2: Simplifying the Equation

Let’s denote:

\[ x = \frac{\cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2}{\cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2}, \]
\[ y = \frac{\cos \theta_3 \cos \theta_4 - \sin \theta_3 \sin \theta_4}{\cos \theta_3 \cos \theta_4 + \sin \theta_3 \sin \theta_4}. \]

Thus, we rewrite our equation as:

\[ x + y = 0 \implies y = -x. \]

### Step 3: Expressing in Terms of Tangents

Now, we convert \(x\) and \(y\) into terms of tangent:

\[ x = \frac{1 + \tan \theta_1 \tan \theta_2}{1 - \tan \theta_1 \tan \theta_2}, \]
\[ y = \frac{1 - \tan \theta_3 \tan \theta_4}{1 + \tan \theta_3 \tan \theta_4}. \]

### Step 4: Setting Up the Equation

Substituting these into our equation gives:

\[ \frac{1 + \tan \theta_1 \tan \theta_2}{1 - \tan \theta_1 \tan \theta_2} + \frac{1 - \tan \theta_3 \tan \theta_4}{1 + \tan \theta_3 \tan \theta_4} = 0. \]

### Step 5: Finding a Common Denominator

To combine these fractions, we find a common denominator:

\[ (1 - \tan \theta_1 \tan \theta_2)(1 + \tan \theta_3 \tan \theta_4). \]

This leads to:

\[ (1 + \tan \theta_1 \tan \theta_2)(1 + \tan \theta_3 \tan \theta_4) + (1 - \tan \theta_3 \tan \theta_4)(1 - \tan \theta_1 \tan \theta_2) = 0. \]

### Step 6: Expanding and Simplifying

Expanding both sides:

\[ 1 + \tan \theta_1 \tan \theta_4 + \tan \theta_2 \tan \theta_3 + \tan \theta_1 \tan \theta_2 \tan \theta_3 \tan \theta_4 + 1 - \tan \theta_1 \tan \theta_2 - \tan \theta_3 \tan \theta_4 = 0. \]

Combining like terms gives:

\[ 2 + \tan \theta_1 \tan \theta_2 \tan \theta_3 \tan \theta_4 - \tan \theta_1 \tan \theta_2 - \tan \theta_3 \tan \theta_4 = 0. \]

### Step 7: Rearranging the Equation

Rearranging leads us to:

\[ \tan \theta_1 \tan \theta_2 \tan \theta_3 \tan \theta_4 = -1. \]

### Final Answer

Thus, we find that:

\[ \tan \theta_1 \tan \theta_2 \tan \theta_3 \tan \theta_4 = -1. \]