From algebra we know that if `ax^(2) +bx + c=0 , a( ne 0), b, c 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 value of `cos(theta_(1)-theta_(2))` is (a and b not being simultaneously zero)

Options

A. (a)`-1 (2c^(2))/(a^(2)+b^(2))`
B. (b)`(2c)/(a^(2)+b^(2))+1`
C. (c)`1-(2c^(2))/(a^(2)+b^(2))`
D. (d)`1-(2c^(2))/(a^(2)+b^(2))`

Answer

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

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

### Step 1: Rearranging the equation
We can rearrange the equation to isolate \( a \cos \theta \):

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

### Step 2: Squaring both sides
Now, we square both sides of the equation:

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

This expands to:

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

### Step 3: Using the Pythagorean Identity
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can express \( \cos^2 \theta \) as \( 1 - \sin^2 \theta \):

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

### Step 4: Rearranging the equation
Rearranging gives us:

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

Combining like terms results in:

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

### Step 5: Identifying the quadratic form
This is a quadratic equation in terms of \( \sin \theta \):

\[ A \sin^2 \theta + B \sin \theta + C = 0 \]

where:
- \( A = b^2 + a^2 \)
- \( B = -2bc \)
- \( C = c^2 - a^2 \)

### Step 6: Finding the roots
Using the quadratic formula, the roots \( \sin \theta_1 \) and \( \sin \theta_2 \) can be found as:

\[ \sin \theta_{1,2} = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]

### Step 7: Finding the product and sum of roots
From Vieta's formulas, we know:
- The sum of the roots \( \sin \theta_1 + \sin \theta_2 = -\frac{B}{A} = \frac{2bc}{b^2 + a^2} \)
- The product of the roots \( \sin \theta_1 \sin \theta_2 = \frac{C}{A} = \frac{c^2 - a^2}{b^2 + a^2} \)

### Step 8: Finding \( \cos(\theta_1 - \theta_2) \)
Using the cosine difference formula:

\[ \cos(\theta_1 - \theta_2) = \cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2 \]

We can express \( \cos \theta_1 \) and \( \cos \theta_2 \) in terms of \( \sin \theta_1 \) and \( \sin \theta_2 \):

\[ \cos \theta_1 = \sqrt{1 - \sin^2 \theta_1}, \quad \cos \theta_2 = \sqrt{1 - \sin^2 \theta_2} \]

Substituting these into the cosine difference formula, we can simplify and find:

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

Combining these gives us:

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

### Final Result
Thus, the value of \( \cos(\theta_1 - \theta_2) \) is:

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