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 ` alpha ` is a root (i) and ` beta` is a root of (ii), then ` tan alpha + tan beta ` can be equal to

Options

A. `(1+sqrt(69)//6`
B. `-1-sqrt(69)//6`
C. `(-3 +sqrt(69))/(6)`
D. `(3-sqrt(69))/(3)`

Answer

To solve the given equations and find the value of \( \tan \alpha + \tan \beta \), let's break down the problem step by step.

### Step 1: Solve the first equation

The first equation is:
\[ 5 \sin^2 x + 3 \sin x \cos x - 3 \cos^2 x = 2 \]

Rearranging gives:
\[ 5 \sin^2 x + 3 \sin x \cos x - 3 \cos^2 x - 2 = 0 \]

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

Using the identity \( \sec^2 x = 1 + \tan^2 x \), we can substitute \( \frac{2}{\cos^2 x} \) with \( 2 \sec^2 x \):
\[ 5 \tan^2 x + 3 \tan x - 3 - 2(1 + \tan^2 x) = 0 \]

This simplifies to:
\[ (5 - 2) \tan^2 x + 3 \tan x - 5 = 0 \]
\[ 3 \tan^2 x + 3 \tan x - 5 = 0 \]

### Step 2: Solve the quadratic equation

Now we can solve the quadratic equation:
\[ 3 \tan^2 x + 3 \tan x - 5 = 0 \]

Using the quadratic formula \( \tan x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 3 \), \( b = 3 \), and \( c = -5 \).

Calculating the discriminant:
\[ D = b^2 - 4ac = 3^2 - 4 \cdot 3 \cdot (-5) = 9 + 60 = 69 \]

Now substituting into the formula:
\[ \tan x = \frac{-3 \pm \sqrt{69}}{6} \]

This gives us two potential values for \( \tan \alpha \):
\[ \tan \alpha = \frac{-3 + \sqrt{69}}{6} \quad \text{or} \quad \tan \alpha = \frac{-3 - \sqrt{69}}{6} \]

### 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 \( \sin 2x = 2 \sin x \cos x \):
\[ \sin^2 x - (\cos^2 x - \sin^2 x) = 2 - 2 \sin x \cos x \]
\[ \sin^2 x + \sin^2 x - \cos^2 x = 2 - 2 \sin x \cos x \]
\[ 2 \sin^2 x - \cos^2 x = 2 - 2 \sin x \cos x \]

Dividing by \( \cos^2 x \):
\[ 2 \tan^2 x - 1 = 2 \sec^2 x - 2 \tan x \]

Substituting \( \sec^2 x = 1 + \tan^2 x \):
\[ 2 \tan^2 x - 1 = 2(1 + \tan^2 x) - 2 \tan x \]
\[ 2 \tan^2 x - 1 = 2 + 2 \tan^2 x - 2 \tan x \]

Rearranging gives:
\[ -1 + 2 = 2 \tan x \quad \Rightarrow \quad 1 = 2 \tan x \quad \Rightarrow \quad \tan x = \frac{1}{2} \]

Thus, we have:
\[ \tan \beta = \frac{3}{2} \]

### Step 4: Find \( \tan \alpha + \tan \beta \)

Now we can find \( \tan \alpha + \tan \beta \):
\[ \tan \alpha + \tan \beta = \left(\frac{-3 + \sqrt{69}}{6}\right) + \frac{3}{2} \]

Converting \( \frac{3}{2} \) to have a common denominator of 6:
\[ \tan \alpha + \tan \beta = \frac{-3 + \sqrt{69}}{6} + \frac{9}{6} = \frac{-3 + \sqrt{69} + 9}{6} = \frac{6 + \sqrt{69}}{6} \]

Thus, the final result is:
\[ \tan \alpha + \tan \beta = 1 + \frac{\sqrt{69}}{6} \]