Prove that the following identity is true. sec 0 + 1 tan 0 %3D tan 0 sec 0 – 1
Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE:
1. Give the measures of the complement and the supplement of an angle measuring 35°.
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![**Proving the Trigonometric Identity**
We are tasked with proving the following identity:
\[
\frac{\sec \theta + 1}{\tan \theta} = \frac{\tan \theta}{\sec \theta - 1}
\]
### Step-by-Step Solution:
We begin with the left side of the equation by multiplying the numerator and denominator by the conjugate of the numerator. This allows us to use a Pythagorean identity to simplify the expression.
1. Start by rewriting the left side:
\[
\frac{\sec \theta + 1}{\tan \theta} = \frac{\sec \theta + 1}{\tan \theta} \cdot \frac{\sec \theta - 1}{\sec \theta - 1}
\]
2. Multiply and simplify using the difference of squares:
\[
\frac{(\sec \theta + 1)(\sec \theta - 1)}{\tan \theta (\sec \theta - 1)} = \frac{\sec^2 \theta - 1}{\tan \theta (\sec \theta - 1)}
\]
3. Apply the Pythagorean identity: \(\sec^2 \theta - 1 = \tan^2 \theta\)
\[
= \frac{\tan^2 \theta}{\tan \theta (\sec \theta - 1)}
\]
4. Cancel \(\tan \theta\) from the numerator and the denominator:
\[
= \frac{\tan \theta}{\sec \theta - 1}
\]
Thus, the identity is proven:
\[
\frac{\sec \theta + 1}{\tan \theta} = \frac{\tan \theta}{\sec \theta - 1}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F14714604-619f-4658-81f0-a6b2ccf0a695%2Fd6304904-fec7-4a7b-bbef-8175a601a3d7%2F2pxc2ja_processed.png&w=3840&q=75)
Transcribed Image Text:**Proving the Trigonometric Identity**
We are tasked with proving the following identity:
\[
\frac{\sec \theta + 1}{\tan \theta} = \frac{\tan \theta}{\sec \theta - 1}
\]
### Step-by-Step Solution:
We begin with the left side of the equation by multiplying the numerator and denominator by the conjugate of the numerator. This allows us to use a Pythagorean identity to simplify the expression.
1. Start by rewriting the left side:
\[
\frac{\sec \theta + 1}{\tan \theta} = \frac{\sec \theta + 1}{\tan \theta} \cdot \frac{\sec \theta - 1}{\sec \theta - 1}
\]
2. Multiply and simplify using the difference of squares:
\[
\frac{(\sec \theta + 1)(\sec \theta - 1)}{\tan \theta (\sec \theta - 1)} = \frac{\sec^2 \theta - 1}{\tan \theta (\sec \theta - 1)}
\]
3. Apply the Pythagorean identity: \(\sec^2 \theta - 1 = \tan^2 \theta\)
\[
= \frac{\tan^2 \theta}{\tan \theta (\sec \theta - 1)}
\]
4. Cancel \(\tan \theta\) from the numerator and the denominator:
\[
= \frac{\tan \theta}{\sec \theta - 1}
\]
Thus, the identity is proven:
\[
\frac{\sec \theta + 1}{\tan \theta} = \frac{\tan \theta}{\sec \theta - 1}
\]
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