Find the value of `tan^(-1) (-tan.(13pi)/(8)) + cot^(-1) (-cot((9pi)/(8)))`

Explanation

To solve the expression \( \tan^{-1}(-\tan(13\pi/8)) + \cot^{-1}(-\cot(9\pi/8)) \), we can follow these steps:

### Step 1: Rewrite the Inverse Functions
Using the identities:
- \( \tan^{-1}(-x) = -\tan^{-1}(x) \)
- \( \cot^{-1}(-x) = \pi - \cot^{-1}(x) \)

We can rewrite the expression as:
\[ \tan^{-1}(-\tan(13\pi/8)) + \cot^{-1}(-\cot(9\pi/8)) = -\tan^{-1}(\tan(13\pi/8)) + \left(\pi - \cot^{-1}(\cot(9\pi/8))\right) \]

### Step 2: Simplify \( \tan^{-1}(\tan(13\pi/8)) \)
The angle \( 13\pi/8 \) is outside the principal range of \( \tan^{-1} \) which is \( (-\pi/2, \pi/2) \). To find an equivalent angle within this range, we can express \( 13\pi/8 \) in terms of a coterminal angle:
\[ 13\pi/8 = 2\pi - 3\pi/8 \]
Using the property \( \tan(2\pi - x) = -\tan(x) \), we have:
\[ \tan(13\pi/8) = -\tan(3\pi/8) \]
Thus,
\[ \tan^{-1}(\tan(13\pi/8)) = \tan^{-1}(-\tan(3\pi/8)) = -\tan^{-1}(\tan(3\pi/8)) = -\frac{3\pi}{8} \]

### Step 3: Simplify \( \cot^{-1}(\cot(9\pi/8)) \)
Similarly, \( 9\pi/8 \) is also outside the principal range of \( \cot^{-1} \) which is \( (0, \pi) \). We can express \( 9\pi/8 \) as:
\[ 9\pi/8 = \pi + \pi/8 \]
Using the property \( \cot(\pi + x) = \cot(x) \), we have:
\[ \cot(9\pi/8) = \cot(\pi/8) \]
Thus,
\[ \cot^{-1}(\cot(9\pi/8)) = \cot^{-1}(\cot(\pi/8)) = \frac{\pi}{8} \]

### Step 4: Substitute Back into the Expression
Now substituting back into our expression:
\[ -\tan^{-1}(\tan(13\pi/8)) + \left(\pi - \cot^{-1}(\cot(9\pi/8))\right) = -\left(-\frac{3\pi}{8}\right) + \left(\pi - \frac{\pi}{8}\right) \]
This simplifies to:
\[ \frac{3\pi}{8} + \left(\pi - \frac{\pi}{8}\right) = \frac{3\pi}{8} + \frac{8\pi}{8} - \frac{\pi}{8} = \frac{3\pi}{8} + \frac{7\pi}{8} = \frac{10\pi}{8} = \frac{5\pi}{4} \]

### Step 5: Final Expression
Thus, the value of the given expression is:
\[ \boxed{\pi} \]