Consider the equation `5 sin^2 x + 3 sin x cos x - 3 cos^2 x =2 `.......... (i)
`sin^2 x - cos 2 x =2-sin 2 x `........... (ii)
If ` tan alpha , tan beta ` satisfy (i) and ` cos gamma , cos delta ` satisfy (ii) , then ` tan alpha * tan beta + cos gamma + cos delta ` can be equal to

Options

A. a) `-1`
B. b) `-(5)/(3)+(2)/(sqrt(13))`
C. c) `(5)/(3)-(2)/(sqrt(13)) `
D. d) `(5)/(3)+(2)/(sqrt(13))`

Answer

To solve the given problem step-by-step, we will analyze the two equations provided and derive the required expression.

### Step 1: Solve the first equation
The first equation is:
\[ 5 \sin^2 x + 3 \sin x \cos x - 3 \cos^2 x = 2 \]

We can rearrange this equation:
\[ 5 \sin^2 x + 3 \sin x \cos x - 3 \cos^2 x - 2 = 0 \]

Next, we divide the entire equation by \( \cos^2 x \):
\[ 5 \tan^2 x + 3 \tan x - 3 - 2 \sec^2 x = 0 \]

Using the identity \( \sec^2 x = 1 + \tan^2 x \), we can substitute:
\[ 5 \tan^2 x + 3 \tan x - 3 - 2(1 + \tan^2 x) = 0 \]
This simplifies to:
\[ 5 \tan^2 x + 3 \tan x - 3 - 2 - 2 \tan^2 x = 0 \]
\[ (5 - 2) \tan^2 x + 3 \tan x - 5 = 0 \]
\[ 3 \tan^2 x + 3 \tan x - 5 = 0 \]

Let \( t = \tan x \):
\[ 3t^2 + 3t - 5 = 0 \]

### Step 2: Solve the quadratic equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 3, b = 3, c = -5 \):
\[ t = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 3 \cdot (-5)}}{2 \cdot 3} \]
\[ t = \frac{-3 \pm \sqrt{9 + 60}}{6} \]
\[ t = \frac{-3 \pm \sqrt{69}}{6} \]

Thus, we have:
\[ \tan \alpha = \frac{-3 + \sqrt{69}}{6}, \quad \tan \beta = \frac{-3 - \sqrt{69}}{6} \]

### Step 3: Solve the second equation
The second equation is:
\[ \sin^2 x - \cos 2x = 2 - \sin 2x \]

Using the identities \( \cos 2x = \cos^2 x - \sin^2 x \) and \( \sin 2x = 2 \sin x \cos x \):
\[ \sin^2 x - (\cos^2 x - \sin^2 x) = 2 - 2 \sin x \cos x \]
This simplifies to:
\[ 2 \sin^2 x + \cos^2 x = 2 - 2 \sin x \cos x \]
Using \( \cos^2 x = 1 - \sin^2 x \):
\[ 2 \sin^2 x + (1 - \sin^2 x) = 2 - 2 \sin x \cos x \]
\[ 2 \sin^2 x + 1 - \sin^2 x = 2 - 2 \sin x \cos x \]
\[ \sin^2 x + 2 \sin x \cos x - 1 = 0 \]

### Step 4: Solve for \( \cos x \)
Using the identity \( \sin^2 x = 1 - \cos^2 x \):
\[ 1 - \cos^2 x + 2 \sin x \cos x - 1 = 0 \]
\[ -\cos^2 x + 2 \sin x \cos x = 0 \]
Factoring out \( \cos x \):
\[ \cos x (2 \sin x - \cos x) = 0 \]

Thus, \( \cos x = 0 \) or \( 2 \sin x - \cos x = 0 \).

### Step 5: Find values of \( \cos \gamma \) and \( \cos \delta \)
If \( \cos x = 0 \), then \( \cos \gamma = 0 \) or \( \cos \delta = 0 \).
If \( 2 \sin x - \cos x = 0 \), then \( \sin x = \frac{1}{2} \cos x \).

### Step 6: Calculate the final expression
We need to find:
\[ \tan \alpha \tan \beta + \cos \gamma + \cos \delta \]

Calculating \( \tan \alpha \tan \beta \):
\[ \tan \alpha \tan \beta = \left(\frac{-3 + \sqrt{69}}{6}\right)\left(\frac{-3 - \sqrt{69}}{6}\right) = \frac{(-3)^2 - (\sqrt{69})^2}{36} = \frac{9 - 69}{36} = \frac{-60}{36} = -\frac{5}{3} \]

Assuming \( \cos \gamma = 0 \) and \( \cos \delta = 0 \):
\[ \tan \alpha \tan \beta + \cos \gamma + \cos \delta = -\frac{5}{3} + 0 + 0 = -\frac{5}{3} \]

### Final Answer
Thus, the final value is:
\[ \boxed{-\frac{5}{3}} \]