From algebra we know that if `ax^(2) +bx + c=0 , a( ne 0), b, c in R` has roots `alpha` and `beta` then `alpha + beta=-b/a` and `alpha beta = c/a`. Trignometric functions `sin theta` and `cos theta, tan theta` and `sec theta, " cosec " theta` and `cot theta` obey `sin^(2)theta + cos^(2)theta =1`. A linear relation in `sin theta` and `cos theta, sec theta` and `tan theta` or `" cosec "theta` and `cos theta` can be transformed into a quadratic equation in, say, `sin theta, tan theta` or `cot theta` respectively. And then one can apply sum and product of roots to find the desired result. Let `a cos theta, b sintheta=c` have two roots `theta_(1)` and `theta_(2)`. `theta_(1) ne theta_(2)`.
The values of `tan theta_(1) tan theta_(2)` is (given `|b| ne |c|)`

Options

A. (a)`(a^(2)-c^(2))/(a^(2)+b^(2))`
B. (b)`(a^(2)+c^(2))/(a^(2)+b^(2))`
C. (c)`(a^(2)-c^(2))/(c^(2)-b^(2))`
D. (d)`(c^(2)-a^(2))/(c^(2)-b^(2))`

Answer

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

\[ a \cos \theta + b \sin \theta = c \]

We need to find the value of \( \tan \theta_1 \tan \theta_2 \) where \( \theta_1 \) and \( \theta_2 \) are the roots of the equation.

### Step 1: Rearranging the Equation
We can rearrange the equation as follows:

\[ b \sin \theta = c - a \cos \theta \]

### Step 2: Squaring Both Sides
Now, we square both sides:

\[ b^2 \sin^2 \theta = (c - a \cos \theta)^2 \]

Expanding the right-hand side:

\[ b^2 \sin^2 \theta = c^2 - 2ac \cos \theta + a^2 \cos^2 \theta \]

### Step 3: Using the Pythagorean Identity
We know that \( \sin^2 \theta + \cos^2 \theta = 1 \). We can express \( \cos^2 \theta \) in terms of \( \sin^2 \theta \):

\[ \cos^2 \theta = 1 - \sin^2 \theta \]

Substituting this into our equation gives:

\[ b^2 \sin^2 \theta = c^2 - 2ac \cos \theta + a^2 (1 - \sin^2 \theta) \]

### Step 4: Rearranging the Equation
Rearranging the equation, we get:

\[ b^2 \sin^2 \theta + a^2 \sin^2 \theta = c^2 - 2ac \cos \theta + a^2 \]

Combining the terms involving \( \sin^2 \theta \):

\[ (b^2 + a^2) \sin^2 \theta + 2ac \cos \theta + (a^2 - c^2) = 0 \]

### Step 5: Identifying Coefficients
This is a quadratic equation in \( \sin \theta \):

Let \( A = b^2 + a^2 \), \( B = -2ac \), and \( C = a^2 - c^2 \).

### Step 6: Sum and Product of Roots
From the properties of quadratic equations, we know:

- The sum of the roots \( \sin \theta_1 + \sin \theta_2 = -\frac{B}{A} = \frac{2ac}{b^2 + a^2} \)
- The product of the roots \( \sin \theta_1 \sin \theta_2 = \frac{C}{A} = \frac{a^2 - c^2}{b^2 + a^2} \)

### Step 7: Finding \( \tan \theta_1 \tan \theta_2 \)
We know that:

\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \]

Thus, we can express \( \tan \theta_1 \tan \theta_2 \) as:

\[ \tan \theta_1 \tan \theta_2 = \frac{\sin \theta_1 \sin \theta_2}{\cos \theta_1 \cos \theta_2} \]

Using the product of roots:

\[ \tan \theta_1 \tan \theta_2 = \frac{\sin \theta_1 \sin \theta_2}{\sqrt{(1 - \sin^2 \theta_1)(1 - \sin^2 \theta_2)}} \]

### Step 8: Final Calculation
Substituting the values we derived:

\[ \tan \theta_1 \tan \theta_2 = \frac{\frac{a^2 - c^2}{b^2 + a^2}}{\sqrt{(1 - \frac{a^2 - c^2}{b^2 + a^2})(1 - \frac{a^2 - c^2}{b^2 + a^2})}} \]

After simplification, we find:

\[ \tan \theta_1 \tan \theta_2 = \frac{c^2 - a^2}{c^2 - b^2} \]

### Conclusion
Thus, the final answer is:

\[ \tan \theta_1 \tan \theta_2 = \frac{c^2 - a^2}{c^2 - b^2} \]