3) Show that 1 + cot Pe=csc²e. CSC

Calculus: Early Transcendentals
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Chapter1: Functions And Models
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Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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The image contains the following mathematical expression, posed as a problem for demonstration:

3) Show that \(1 + \cot^2 \theta = \csc^2 \theta\).

This is a trigonometric identity that can be proved using fundamental trigonometric relationships. The identity is derived from the Pythagorean identity for sine and cosine, and involves expressing cotangent and cosecant in terms of sine and cosine functions.

To prove this identity, follow these steps:

1. Recall that \(\cot \theta = \frac{\cos \theta}{\sin \theta}\) and \(\csc \theta = \frac{1}{\sin \theta}\).

2. Substitute these definitions into the identity:  

   \[
   1 + \left( \frac{\cos \theta}{\sin \theta} \right)^2 = \left( \frac{1}{\sin \theta} \right)^2
   \]

3. Simplify the left side:

   \[
   1 + \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta}
   \]

4. Using the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\), the expression becomes:

   \[
   \frac{1}{\sin^2 \theta} = \csc^2 \theta
   \]

Both sides of the equation are now equal, confirming the identity.
Transcribed Image Text:The image contains the following mathematical expression, posed as a problem for demonstration: 3) Show that \(1 + \cot^2 \theta = \csc^2 \theta\). This is a trigonometric identity that can be proved using fundamental trigonometric relationships. The identity is derived from the Pythagorean identity for sine and cosine, and involves expressing cotangent and cosecant in terms of sine and cosine functions. To prove this identity, follow these steps: 1. Recall that \(\cot \theta = \frac{\cos \theta}{\sin \theta}\) and \(\csc \theta = \frac{1}{\sin \theta}\). 2. Substitute these definitions into the identity: \[ 1 + \left( \frac{\cos \theta}{\sin \theta} \right)^2 = \left( \frac{1}{\sin \theta} \right)^2 \] 3. Simplify the left side: \[ 1 + \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta} \] 4. Using the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\), the expression becomes: \[ \frac{1}{\sin^2 \theta} = \csc^2 \theta \] Both sides of the equation are now equal, confirming the identity.
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