The value of `sin^(-1)sin17+cos^(-1)cos10` is equal to

Options

A. 27
B. `-27`
C. `17-5pi`
D. `9pi-27`

Answer

To solve the expression \( \sin^{-1}(\sin 17) + \cos^{-1}(\cos 10) \), we will follow these steps:

### Step 1: Simplify \( \sin^{-1}(\sin 17) \)

The function \( \sin^{-1}(x) \) gives the angle whose sine is \( x \). The range of \( \sin^{-1}(x) \) is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \). Since \( 17 \) is outside this range, we need to find an equivalent angle within the range.

To find an equivalent angle, we can use the property:
\[ \sin^{-1}(\sin x) = x \quad \text{if } x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \]
For \( x = 17 \), we can find an equivalent angle by subtracting \( 2\pi \) or using \( \pi - x \):
\[ 17 - 2\pi \approx 17 - 6.2832 \approx 10.7168 \quad (\text{still outside the range}) \]
Next, we can use:
\[ \sin^{-1}(\sin x) = \pi - x \quad \text{if } x \in \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \]
Since \( 17 \) is approximately \( 5.4\pi \), we can find:
\[ \sin^{-1}(\sin 17) = \pi - (17 - 2\pi) = 3\pi - 17 \]

### Step 2: Simplify \( \cos^{-1}(\cos 10) \)

The function \( \cos^{-1}(x) \) gives the angle whose cosine is \( x \). The range of \( \cos^{-1}(x) \) is \( [0, \pi] \). Since \( 10 \) is outside this range, we can find an equivalent angle using:
\[ \cos^{-1}(\cos x) = x \quad \text{if } x \in [0, \pi] \]
For \( x = 10 \), we can find an equivalent angle:
\[ 10 - 2\pi \approx 10 - 6.2832 \approx 3.7168 \quad (\text{still outside the range}) \]
Next, we can use:
\[ \cos^{-1}(\cos x) = 2\pi - x \quad \text{if } x \in (0, 2\pi) \]
Since \( 10 \) is between \( \pi \) and \( 2\pi \):
\[ \cos^{-1}(\cos 10) = 2\pi - 10 \]

### Step 3: Combine the results

Now we combine the results from Step 1 and Step 2:
\[ \sin^{-1}(\sin 17) + \cos^{-1}(\cos 10) = (3\pi - 17) + (2\pi - 10) \]
\[ = 5\pi - 27 \]

### Final Result

Thus, the value of \( \sin^{-1}(\sin 17) + \cos^{-1}(\cos 10) \) is:
\[ 5\pi - 27 \]