Let's solve this step-by-step.

The problem is:

Given: $a = \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) + \cos^{-1}\left(-\frac{1}{2}\right)$ and $b = \tan^{-1}(\sqrt{3}) - \cot^{-1}\left(-\frac{1}{\sqrt{3}}\right)$

We need to find the value of $a + b$.

Step 1: Calculate $a$

1. Evaluate $\sin^{-1}\left(\frac{\sqrt{2}}{2}\right)$: $\sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}$ because $\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$.

2. Evaluate $\cos^{-1}\left(-\frac{1}{2}\right)$: $\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$ because $\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}$.

Thus, $a = \frac{\pi}{4} + \frac{2\pi}{3}$

To add these, find a common denominator: $a = \frac{3\pi}{12} + \frac{8\pi}{12} = \frac{11\pi}{12}$

Step 2: Calculate $b$

1. Evaluate $\tan^{-1}(\sqrt{3})$: $\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$ because $\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$.

2. Evaluate $\cot^{-1}\left(-\frac{1}{\sqrt{3}}\right)$: We know that $\cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right)$. So: $\cot^{-1}\left(-\frac{1}{\sqrt{3}}\right) = \tan^{-1}(-\sqrt{3})$ and $\tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}$

Thus, $b = \frac{\pi}{3} - \left(-\frac{\pi}{3}\right) = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3}$

Step 3: Calculate $a + b$

Now we have: $a = \frac{11\pi}{12} \quad \text{and} \quad b = \frac{2\pi}{3} = \frac{8\pi}{12}$

So, $a + b = \frac{11\pi}{12} + \frac{8\pi}{12} = \frac{19\pi}{12}$

Final Answer

$a + b = \frac{19\pi}{12}$

Would you like further details or have any questions?

1. What are the properties of inverse trigonometric functions?
2. How can we simplify expressions involving inverse trigonometric functions?
3. Why do we convert inverse cotangent to an inverse tangent?
4. How do we find a common denominator for adding fractions?
5. Can this solution be verified with approximate numerical values?

Tip: Remember that inverse trigonometric functions have specific ranges, which helps in evaluating them accurately.