The value of `tan((sin^(-1)x+cos^(-1)x)/(2))`, when `x = (sqrt(3))/(2)`, is `"……."`.

Explanation

To solve the problem of finding the value of \( \tan\left(\frac{\sin^{-1}x + \cos^{-1}x}{2}\right) \) when \( x = \frac{\sqrt{3}}{2} \), we can follow these steps:

### Step 1: Identify the values of \( \sin^{-1}x \) and \( \cos^{-1}x \)
Given \( x = \frac{\sqrt{3}}{2} \):
- \( \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} \) (since sine of \( \frac{\pi}{3} \) is \( \frac{\sqrt{3}}{2} \))
- \( \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6} \) (since cosine of \( \frac{\pi}{6} \) is \( \frac{\sqrt{3}}{2} \))

### Step 2: Add the values of \( \sin^{-1}x \) and \( \cos^{-1}x \)
Now, we add these two values:
\[ \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) + \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} + \frac{\pi}{6} \]

To add these fractions, we need a common denominator:
- The LCM of 3 and 6 is 6.
- Convert \( \frac{\pi}{3} \) to sixths: \( \frac{\pi}{3} = \frac{2\pi}{6} \)

Now, we can add:
\[ \frac{2\pi}{6} + \frac{\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2} \]

### Step 3: Substitute back into the original expression
Now, we substitute this result back into the original function:
\[ \tan\left(\frac{\sin^{-1}x + \cos^{-1}x}{2}\right) = \tan\left(\frac{\frac{\pi}{2}}{2}\right) = \tan\left(\frac{\pi}{4}\right) \]

### Step 4: Calculate the final value
We know that:
\[ \tan\left(\frac{\pi}{4}\right) = 1 \]

Thus, the final answer is:
\[ \boxed{1} \]