The value of $\tan^{-1}(1) + \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) + \sin^{-1}\left(\frac{1}{2}\right)$ is

The angle with tangent value of 1 occurs when opposite = adjacent, which happens at 45° or $\frac{\pi}{4}$ radians.

$\tan^{-1}(1) = \frac{\pi}{4}$

For cosine value of $-\frac{\sqrt{3}}{2}$, the negative sign indicates the 2nd or 3rd quadrant. The range of inverse cosine is $[0, \pi]$, so the 2nd quadrant is used.

Since $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$

In the 2nd quadrant:

$\pi - \frac{\pi}{6} = \frac{5\pi}{6}$

$\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) = \frac{5\pi}{6}$

The sine value of $\frac{1}{2}$ is a standard value:

$\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$

$\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$

Adding all three values:

$\frac{\pi}{4} + \frac{5\pi}{6} + \frac{\pi}{6}$

Converting to common denominator of 12:

$\frac{3\pi}{12} + \frac{10\pi}{12} + \frac{2\pi}{12}$

$= \frac{15\pi}{12}$

$= \frac{5\pi}{4}$

Therefore, the value is $\frac{5\pi}{4}$.