Prove that `1+cot^(2) theta = cosec^(2) theta`

Answer

To prove that \( 1 + \cot^2 \theta = \csc^2 \theta \), we will use the definitions of trigonometric ratios and the Pythagorean theorem.

### Step-by-Step Solution:

1. **Understanding the Definitions**:
- In a right triangle, let \( \theta \) be one of the angles.
- The sides of the triangle are defined as follows:
- Opposite side to \( \theta \): \( P \) (Perpendicular)
- Adjacent side to \( \theta \): \( B \) (Base)
- Hypotenuse: \( H \)

2. **Write the Definitions of Cotangent and Cosecant**:
- The cotangent of \( \theta \) is defined as:
\[ \cot \theta = \frac{B}{P} \]
- The cosecant of \( \theta \) is defined as:
\[ \csc \theta = \frac{H}{P} \]

3. **Applying the Pythagorean Theorem**:
- According to the Pythagorean theorem, we have:
\[ H^2 = P^2 + B^2 \]

4. **Dividing the Entire Equation by \( P^2 \)**:
- We divide the equation \( H^2 = P^2 + B^2 \) by \( P^2 \):
\[ \frac{H^2}{P^2} = \frac{P^2}{P^2} + \frac{B^2}{P^2} \]
- This simplifies to:
\[ \frac{H^2}{P^2} = 1 + \frac{B^2}{P^2} \]

5. **Substituting the Trigonometric Ratios**:
- We know that:
\[ \frac{H}{P} = \csc \theta \quad \text{and} \quad \frac{B}{P} = \cot \theta \]
- Therefore, we can rewrite the equation as:
\[ \left(\csc \theta\right)^2 = 1 + \left(\cot \theta\right)^2 \]

6. **Final Result**:
- Rearranging gives us:
\[ 1 + \cot^2 \theta = \csc^2 \theta \]
- This completes the proof.