It f and g be function defined by `f(theta) = cos^2 theta and(theta) = tan^2 theta,` suppose `alpha and beta` satisfy `2 f(alpha) – g(beta) =1,`then value of `2 f(beta) – g(alpha)` is

Answer

To solve the problem, we need to evaluate the expression \(2f(\beta) - g(\alpha)\) given the functions \(f(\theta) = \cos^2 \theta\) and \(g(\theta) = \tan^2 \theta\), and the condition \(2f(\alpha) - g(\beta) = 1\).

### Step-by-step Solution:

1. **Define the functions**:
\[ f(\theta) = \cos^2 \theta \]
\[ g(\theta) = \tan^2 \theta \]

2. **Substitute the functions into the given condition**:
The condition is:
\[ 2f(\alpha) - g(\beta) = 1 \]
Substituting the definitions of \(f\) and \(g\):
\[ 2\cos^2 \alpha - \tan^2 \beta = 1 \]

3. **Express \(\tan^2 \beta\) in terms of \(\cos^2 \alpha\)**:
Rearranging the equation gives:
\[ \tan^2 \beta = 2\cos^2 \alpha - 1 \]

4. **Find \(2f(\beta) - g(\alpha)\)**:
We need to evaluate:
\[ 2f(\beta) - g(\alpha) = 2\cos^2 \beta - \tan^2 \alpha \]

5. **Substitute \(\tan^2 \alpha\)**:
We know that:
\[ \tan^2 \alpha = \frac{\sin^2 \alpha}{\cos^2 \alpha} = \frac{1 - \cos^2 \alpha}{\cos^2 \alpha} \]
Thus, we can express \(g(\alpha)\) as:
\[ g(\alpha) = \tan^2 \alpha = \frac{1 - \cos^2 \alpha}{\cos^2 \alpha} \]

6. **Substituting values**:
Let's assume \(\alpha = 0\) and \(\beta = 45^\circ\):
- For \(\alpha = 0\):
\[ f(0) = \cos^2(0) = 1 \]
\[ g(0) = \tan^2(0) = 0 \]
- For \(\beta = 45^\circ\):
\[ f(45^\circ) = \cos^2(45^\circ) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]
\[ g(45^\circ) = \tan^2(45^\circ) = 1 \]

7. **Calculate \(2f(\beta) - g(\alpha)\)**:
Now substituting these values:
\[ 2f(45^\circ) - g(0) = 2 \times \frac{1}{2} - 0 = 1 - 0 = 1 \]

### Final Result:
Thus, the value of \(2f(\beta) - g(\alpha)\) is:
\[ \boxed{1} \]