To verify the given trigonometric identity [tex]\( 1 + \cot^2 \theta = \csc^2 \theta \)[/tex], let's go step by step:
1. Understand the trigonometric functions involved:
- [tex]\(\cot \theta\)[/tex] is the cotangent of [tex]\(\theta\)[/tex], which is defined as [tex]\(\frac{\cos \theta}{\sin \theta}\)[/tex].
- [tex]\(\csc \theta\)[/tex] is the cosecant of [tex]\(\theta\)[/tex], which is defined as [tex]\(\frac{1}{\sin \theta}\)[/tex].
2. Rewrite the left-hand side (LHS) of the identity:
- LHS: [tex]\(1 + \cot^2 \theta\)[/tex]
- Substitute [tex]\(\cot \theta = \frac{\cos \theta}{\sin \theta}\)[/tex]:
[tex]\[
\cot^2 \theta = \left(\frac{\cos \theta}{\sin \theta}\right)^2 = \frac{\cos^2 \theta}{\sin^2 \theta}
\][/tex]
- Therefore, the LHS becomes:
[tex]\[
1 + \cot^2 \theta = 1 + \frac{\cos^2 \theta}{\sin^2 \theta}
\][/tex]
3. Convert the LHS to a common denominator:
- Express 1 as [tex]\(\frac{\sin^2 \theta}{\sin^2 \theta}\)[/tex] to have a common denominator:
[tex]\[
1 + \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{\sin^2 \theta}{\sin^2 \theta} + \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta}
\][/tex]
4. Use the Pythagorean identity:
- Recall the Pythagorean identity: [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex]
- Substitute this into the LHS:
[tex]\[
\frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta} = \frac{1}{\sin^2 \theta}
\][/tex]
5. Rewrite the expression using cosecant:
- We know that [tex]\(\csc \theta = \frac{1}{\sin \theta}\)[/tex]. Therefore:
[tex]\[
\frac{1}{\sin^2 \theta} = \csc^2 \theta
\][/tex]
6. Combine all steps to conclude:
- The LHS [tex]\(1 + \cot^2 \theta\)[/tex] simplifies to [tex]\(\csc^2 \theta\)[/tex].
- Hence, we have shown that [tex]\(1 + \cot^2 \theta = \csc^2 \theta\)[/tex].
This verifies the given trigonometric identity.
