If `tan theta.tan phi = sqrt ((a -b)/(a +b)), ` then `(a -b cos 2 theta)(a -b cos 2 phi)` is

Options

A. independent of `theta`
B. independent of `phi`
C. independent of `theta and phi`
D. none of these

Answer

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

\[ \tan \theta \tan \phi = \sqrt{\frac{a - b}{a + b}} \]

### Step 1: Square both sides
We square both sides of the equation to eliminate the square root:

\[ (\tan \theta \tan \phi)^2 = \frac{a - b}{a + b} \]

Let \( t_1 = \tan \theta \) and \( t_2 = \tan \phi \). Thus, we can rewrite the equation as:

\[ t_1^2 t_2^2 = \frac{a - b}{a + b} \]

### Step 2: Use the double angle formula for cosine
We know that:

\[ \cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \frac{1 - t_1^2}{1 + t_1^2} \]

And similarly for \(\phi\):

\[ \cos 2\phi = \frac{1 - t_2^2}{1 + t_2^2} \]

### Step 3: Substitute into the expression
We need to find the expression \( (a - b \cos 2\theta)(a - b \cos 2\phi) \).

Substituting the values of \(\cos 2\theta\) and \(\cos 2\phi\):

\[ a - b \cos 2\theta = a - b \left( \frac{1 - t_1^2}{1 + t_1^2} \right) = a - b \frac{1 - t_1^2}{1 + t_1^2} \]

This simplifies to:

\[ = \frac{(a(1 + t_1^2) - b(1 - t_1^2))}{1 + t_1^2} = \frac{(a + at_1^2 - b + bt_1^2)}{1 + t_1^2} = \frac{(a - b) + (a + b)t_1^2}{1 + t_1^2} \]

Similarly, for \(\phi\):

\[ a - b \cos 2\phi = \frac{(a - b) + (a + b)t_2^2}{1 + t_2^2} \]

### Step 4: Multiply the two expressions
Now we multiply the two results:

\[ (a - b \cos 2\theta)(a - b \cos 2\phi) = \left( \frac{(a - b) + (a + b)t_1^2}{1 + t_1^2} \right) \left( \frac{(a - b) + (a + b)t_2^2}{1 + t_2^2} \right) \]

### Step 5: Simplify the product
The product can be simplified as follows:

\[ = \frac{((a - b) + (a + b)t_1^2)((a - b) + (a + b)t_2^2)}{(1 + t_1^2)(1 + t_2^2)} \]

### Step 6: Substitute \(t_1^2 t_2^2\)
From our earlier step, we know:

\[ t_1^2 t_2^2 = \frac{a - b}{a + b} \]

Thus, we can substitute this into our expression. The numerator becomes:

\[ (a - b)^2 + (a + b)(t_1^2 + t_2^2)(a - b) + (a + b)^2 t_1^2 t_2^2 \]

### Step 7: Final expression
After simplification, we find that:

\[ (a - b \cos 2\theta)(a - b \cos 2\phi) = \frac{(a - b)(a + b)}{(1 + t_1^2)(1 + t_2^2)} \]

### Conclusion
Thus, the final result is:

\[ (a - b \cos 2\theta)(a - b \cos 2\phi) = \frac{(a - b)(a + b)}{(1 + t_1^2)(1 + t_2^2)} \]