`sin^(2)alpha+cos^(2)beta=2` then `tan((alpha+beta)/(2))=?`

Answer

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

**Step 1: Analyze the equation.**
We have:
\[ \sin^2 \alpha + \cos^2 \beta = 2 \]

**Step 2: Identify the maximum values of sine and cosine.**
The maximum value of \(\sin^2 \alpha\) is 1 and the maximum value of \(\cos^2 \beta\) is also 1. Therefore, for the sum \(\sin^2 \alpha + \cos^2 \beta\) to equal 2, both terms must be at their maximum values:
\[ \sin^2 \alpha = 1 \quad \text{and} \quad \cos^2 \beta = 1 \]

**Step 3: Solve for \(\alpha\) and \(\beta\).**
From \(\sin^2 \alpha = 1\), we take the square root:
\[ \sin \alpha = 1 \]
This occurs when:
\[ \alpha = 90^\circ \]

From \(\cos^2 \beta = 1\), we take the square root:
\[ \cos \beta = 1 \]
This occurs when:
\[ \beta = 0^\circ \]

**Step 4: Find \(\tan\left(\frac{\alpha + \beta}{2}\right)\).**
Now we need to calculate:
\[ \tan\left(\frac{\alpha + \beta}{2}\right) = \tan\left(\frac{90^\circ + 0^\circ}{2}\right) = \tan\left(\frac{90^\circ}{2}\right) = \tan(45^\circ) \]

**Step 5: Calculate \(\tan(45^\circ)\).**
We know that:
\[ \tan(45^\circ) = 1 \]

Thus, the final answer is:
\[ \tan\left(\frac{\alpha + \beta}{2}\right) = 1 \]

---