To solve the given mathematical expression [tex]\(Z = \sqrt{3} + \sqrt{-\cot \frac{25 \pi}{4}}\)[/tex], we need to follow a clear, step-by-step approach:
1. Evaluate the angle [tex]\( \frac{25 \pi}{4} \)[/tex]:
Note that angles in trigonometric functions can be converted to a corresponding angle within a standard period by subtracting or adding multiples of [tex]\(2\pi\)[/tex] (or [tex]\(360^\circ\)[/tex]).
[tex]\[
\frac{25 \pi}{4} = 2\pi \cdot 6 + \frac{\pi}{4}
\][/tex]
Hence,
[tex]\[
\frac{25 \pi}{4} \equiv \frac{25 \pi}{4} - 2\pi \cdot 6 = \frac{\pi}{4} \ (\text{mod}\ 2\pi)
\][/tex]
Therefore:
[tex]\[
\frac{25 \pi}{4} \equiv \frac{\pi}{4}
\][/tex]
2. Evaluate [tex]\( \cot \frac{\pi}{4} \)[/tex]:
The cotangent function [tex]\( \cot \theta \)[/tex] is defined as the reciprocal of the tangent function:
[tex]\[
\cot \theta = \frac{1}{\tan \theta}
\][/tex]
For [tex]\( \frac{\pi}{4} \)[/tex]:
[tex]\[
\tan \frac{\pi}{4} = 1 \quad \text{(since the tangent of \(45^\circ\) is 1)}
\][/tex]
Therefore,
[tex]\[
\cot \frac{\pi}{4} = \frac{1}{\tan \frac{\pi}{4}} = \frac{1}{1} = 1
\][/tex]
3. Evaluate [tex]\( -\cot \frac{\pi}{4} \)[/tex]:
[tex]\[
-\cot \frac{\pi}{4} = -1
\][/tex]
So the expression becomes:
[tex]\[
Z = \sqrt{3} + \sqrt{-1}
\][/tex]
4. Simplify [tex]\( \sqrt{-1} \)[/tex]:
The square root of [tex]\(-1\)[/tex] is the imaginary unit, denoted by [tex]\(i\)[/tex]:
[tex]\[
\sqrt{-1} = i
\][/tex]
5. Combine the terms:
Finally, substitute this back into the expression for [tex]\(Z\)[/tex]:
[tex]\[
Z = \sqrt{3} + i
\][/tex]
Therefore, the final simplified form of the given expression is:
[tex]\[
\boxed{\sqrt{3} + i}
\][/tex]
