Prove each of the following
`tan^(-1) x=-pi +cot^(-1) 1/x=sin^(-1) (x)/(sqrt(1+x^(2))`
`=-cos^(-1) (1)/(sqrt(1+x^(2))" when "x lt 0`

Explanation

To prove the given equations for \( x < 0 \):

1. **Prove that \( \tan^{-1} x = -\pi + \cot^{-1} \frac{1}{x} \)**

Let's start by letting \( \tan^{-1} x = \theta \). This means that:
\[ \tan \theta = x \]
Since \( \tan \theta = \frac{1}{\cot \theta} \), we can express \( \cot \theta \) as:
\[ \cot \theta = \frac{1}{x} \]
The range of \( \tan^{-1} x \) for \( x < 0 \) is \( (-\frac{\pi}{2}, 0) \). The cotangent function, \( \cot^{-1} \), is defined for \( (0, \pi) \). Therefore, we can express \( \theta \) in terms of \( \cot^{-1} \):
\[ \theta = \cot^{-1} \frac{1}{x} - \pi \]
Thus, we have:
\[ \tan^{-1} x = -\pi + \cot^{-1} \frac{1}{x} \]

2. **Prove that \( \tan^{-1} x = \frac{\sin^{-1} x}{\sqrt{1+x^2}} \)**

From the definition of tangent, we know:
\[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{1} \]
The hypotenuse \( h \) can be calculated using the Pythagorean theorem:
\[ h = \sqrt{x^2 + 1} \]
Therefore, we can find \( \sin \theta \):
\[ \sin \theta = \frac{x}{\sqrt{x^2 + 1}} \]
Thus, we can express \( \theta \) in terms of \( \sin^{-1} \):
\[ \theta = \sin^{-1} \left( \frac{x}{\sqrt{x^2 + 1}} \right) \]
Hence,
\[ \tan^{-1} x = \frac{\sin^{-1} x}{\sqrt{1+x^2}} \]

3. **Prove that \( \tan^{-1} x = -\cos^{-1} \frac{1}{\sqrt{1+x^2}} \)**

From the previous step, we know:
\[ \cos \theta = \frac{1}{h} = \frac{1}{\sqrt{x^2 + 1}} \]
Therefore, we can express \( \theta \) in terms of \( \cos^{-1} \):
\[ \theta = -\cos^{-1} \left( \frac{1}{\sqrt{1+x^2}} \right) \]
Thus,
\[ \tan^{-1} x = -\cos^{-1} \frac{1}{\sqrt{1+x^2}} \]

### Summary of Results:
- \( \tan^{-1} x = -\pi + \cot^{-1} \frac{1}{x} \)
- \( \tan^{-1} x = \frac{\sin^{-1} x}{\sqrt{1+x^2}} \)
- \( \tan^{-1} x = -\cos^{-1} \frac{1}{\sqrt{1+x^2}} \)