If `tan^(2)alpha=2tan^(2)beta+1`, evaluate : `cos 2alpha+sin^(2)beta.`

Answer

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

\[ \tan^2 \alpha = 2 \tan^2 \beta + 1 \]

We need to evaluate:

\[ \cos 2\alpha + \sin^2 \beta \]

### Step 1: Express \(\cos 2\alpha\) in terms of \(\tan^2 \alpha\)

Using the double angle formula for cosine, we have:

\[ \cos 2\alpha = \frac{1 - \tan^2 \alpha}{1 + \tan^2 \alpha} \]

### Step 2: Substitute \(\tan^2 \alpha\)

Substituting the expression for \(\tan^2 \alpha\):

\[ \cos 2\alpha = \frac{1 - (2 \tan^2 \beta + 1)}{1 + (2 \tan^2 \beta + 1)} \]

### Step 3: Simplify the expression

Now simplify the numerator and denominator:

- **Numerator:**
\[ 1 - (2 \tan^2 \beta + 1) = 1 - 2 \tan^2 \beta - 1 = -2 \tan^2 \beta \]

- **Denominator:**
\[ 1 + (2 \tan^2 \beta + 1) = 1 + 2 \tan^2 \beta + 1 = 2 + 2 \tan^2 \beta = 2(1 + \tan^2 \beta) \]

Thus, we have:

\[ \cos 2\alpha = \frac{-2 \tan^2 \beta}{2(1 + \tan^2 \beta)} = \frac{-\tan^2 \beta}{1 + \tan^2 \beta} \]

### Step 4: Use the identity for \(\tan^2 \beta\)

Recall that:

\[ \tan^2 \beta = \frac{\sin^2 \beta}{\cos^2 \beta} \]

Substituting this into our expression for \(\cos 2\alpha\):

\[ \cos 2\alpha = \frac{-\frac{\sin^2 \beta}{\cos^2 \beta}}{1 + \frac{\sin^2 \beta}{\cos^2 \beta}} = \frac{-\sin^2 \beta}{\cos^2 \beta + \sin^2 \beta} = \frac{-\sin^2 \beta}{1} \]

Thus:

\[ \cos 2\alpha = -\sin^2 \beta \]

### Step 5: Combine with \(\sin^2 \beta\)

Now, we can evaluate \(\cos 2\alpha + \sin^2 \beta\):

\[ \cos 2\alpha + \sin^2 \beta = -\sin^2 \beta + \sin^2 \beta = 0 \]

### Final Answer

The value of \(\cos 2\alpha + \sin^2 \beta\) is:

\[ \boxed{0} \]