find the value of `sin^(-1)(-sqrt3/2)+cos^(-1)(sqrt3/2)`

Explanation

To solve the problem \( \sin^{-1}(-\sqrt{3}/2) + \cos^{-1}(\sqrt{3}/2) \), we will follow these steps:

### Step 1: Evaluate \( \sin^{-1}(-\sqrt{3}/2) \)

The value \( \sin^{-1}(-\sqrt{3}/2) \) corresponds to an angle whose sine is \(-\sqrt{3}/2\). The sine function is negative in the fourth quadrant. The reference angle where sine is \(\sqrt{3}/2\) is \(\pi/3\). Therefore, we have:

\[ \sin^{-1}(-\sqrt{3}/2) = -\frac{\pi}{3} \]

### Step 2: Evaluate \( \cos^{-1}(\sqrt{3}/2) \)

The value \( \cos^{-1}(\sqrt{3}/2) \) corresponds to an angle whose cosine is \(\sqrt{3}/2\). The angle in the first quadrant where cosine is \(\sqrt{3}/2\) is \(\pi/6\). Therefore, we have:

\[ \cos^{-1}(\sqrt{3}/2) = \frac{\pi}{6} \]

### Step 3: Combine the results

Now, we add the two results together:

\[ \sin^{-1}(-\sqrt{3}/2) + \cos^{-1}(\sqrt{3}/2) = -\frac{\pi}{3} + \frac{\pi}{6} \]

### Step 4: Find a common denominator

To add these fractions, we need a common denominator. The least common multiple of 3 and 6 is 6. We rewrite \(-\frac{\pi}{3}\) as:

\[ -\frac{\pi}{3} = -\frac{2\pi}{6} \]

Now we can add:

\[ -\frac{2\pi}{6} + \frac{\pi}{6} = -\frac{2\pi - \pi}{6} = -\frac{\pi}{6} \]

### Final Answer

Thus, the final value is:

\[ \sin^{-1}(-\sqrt{3}/2) + \cos^{-1}(\sqrt{3}/2) = -\frac{\pi}{6} \]

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