A continuous, even periodic function f with period 8 is such that `f(0)=0,f(1)=-2,f(2)=1,f(3)=2,f(4)=3,` then the value of `tan^(-1) tan{f(-5)+f(20)+cos^(-1)(f(-10))+f(17)}` is equal to

Options

A. `2pi-3`
B. `3-2pi`
C. `2pi+3`
D. `3-pi`

Explanation

To solve the problem step by step, we will utilize the properties of the periodic and even function \( f \) and the given values.

### Step 1: Understand the periodic function
Since \( f \) is a periodic function with a period of 8, we have:
\[ f(x) = f(x + 8n) \quad \text{for any integer } n \]
This means we can find \( f \) at negative and larger values by reducing them modulo 8.

### Step 2: Calculate \( f(-5) \)
To find \( f(-5) \):
\[ -5 + 8 = 3 \quad \Rightarrow \quad f(-5) = f(3) \]
From the given values, \( f(3) = 2 \). Thus,
\[ f(-5) = 2 \]

### Step 3: Calculate \( f(20) \)
To find \( f(20) \):
\[ 20 - 16 = 4 \quad \Rightarrow \quad f(20) = f(4) \]
From the given values, \( f(4) = 3 \). Thus,
\[ f(20) = 3 \]

### Step 4: Calculate \( f(-10) \)
To find \( f(-10) \):
\[ -10 + 16 = 6 \quad \Rightarrow \quad f(-10) = f(6) \]
Since \( f \) is even, \( f(-2) = f(2) \). We need to find \( f(2) \):
From the given values, \( f(2) = 1 \). Therefore,
\[ f(-10) = f(6) = f(2) = 1 \]

### Step 5: Calculate \( f(17) \)
To find \( f(17) \):
\[ 17 - 16 = 1 \quad \Rightarrow \quad f(17) = f(1) \]
From the given values, \( f(1) = -2 \). Thus,
\[ f(17) = -2 \]

### Step 6: Combine the results
Now we can substitute these values into the expression:
\[ f(-5) + f(20) + \cos^{-1}(f(-10)) + f(17) \]
Substituting the values we found:
\[ = 2 + 3 + \cos^{-1}(1) - 2 \]
Since \( \cos^{-1}(1) = 0 \):
\[ = 2 + 3 + 0 - 2 = 3 \]

### Step 7: Calculate \( \tan^{-1}(\tan(3)) \)
Now we need to evaluate:
\[ \tan^{-1}(\tan(3)) \]
Since \( \tan^{-1}(x) \) returns values in the range \( (-\frac{\pi}{2}, \frac{\pi}{2}) \), and \( 3 \) is outside this range, we need to adjust it:
\[ 3 - \pi \quad \text{(since } 3 \text{ is in the range of } \pi \text{ to } 2\pi\text{)} \]
Thus,
\[ \tan^{-1}(\tan(3)) = 3 - \pi \]

### Final Answer
The value of \( \tan^{-1}(\tan(f(-5) + f(20) + \cos^{-1}(f(-10)) + f(17))) \) is:
\[ \boxed{3 - \pi} \]