`2-cos^(2)theta =3 sintheta costheta , sintheta ne cos theta`, then the value of `tan theta` is

Options

A. `2/3`
B. `1/3`
C. `1/2`
D. 0

Answer

To solve the equation \( 2 - \cos^2 \theta = 3 \sin \theta \cos \theta \) and find the value of \( \tan \theta \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 2 - \cos^2 \theta = 3 \sin \theta \cos \theta \]

### Step 2: Divide by \( \cos^2 \theta \)
To express everything in terms of \( \tan \theta \), we divide the entire equation by \( \cos^2 \theta \):
\[ \frac{2}{\cos^2 \theta} - \frac{\cos^2 \theta}{\cos^2 \theta} = \frac{3 \sin \theta \cos \theta}{\cos^2 \theta} \]
This simplifies to:
\[ 2 \sec^2 \theta - 1 = 3 \tan \theta \]

### Step 3: Use the identity \( \sec^2 \theta = 1 + \tan^2 \theta \)
Now, we can substitute \( \sec^2 \theta \) with \( 1 + \tan^2 \theta \):
\[ 2(1 + \tan^2 \theta) - 1 = 3 \tan \theta \]
This simplifies to:
\[ 2 + 2 \tan^2 \theta - 1 = 3 \tan \theta \]
or:
\[ 2 \tan^2 \theta - 3 \tan \theta + 1 = 0 \]

### Step 4: Solve the quadratic equation
Let \( x = \tan \theta \). The equation becomes:
\[ 2x^2 - 3x + 1 = 0 \]
We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = -3 \), and \( c = 1 \):
\[ x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} \]
Calculating the discriminant:
\[ x = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4} \]

### Step 5: Find the roots
Calculating the two possible values:
1. \( x = \frac{4}{4} = 1 \)
2. \( x = \frac{2}{4} = \frac{1}{2} \)

### Step 6: Conclusion
Since we are given that \( \sin \theta \neq \cos \theta \), we discard \( \tan \theta = 1 \) (which corresponds to \( \sin \theta = \cos \theta \)). Thus, the valid solution is:
\[ \tan \theta = \frac{1}{2} \]