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Establish the identity. \( \begin{array}{l}1+\frac{\tan ^{2} \theta}{\sec \theta+1}=\sec \theta \\ \text { Simplify the numerator of the fraction from the left side by applying the appropriate Pythagorean Identity. } \\ 1+\frac{\sec ^{2} \theta-1}{\sec \theta+1} \text { (Do not factor.) } \\ \text { Factor the numerator of the fraction from the previous step. } \\ 1+\frac{\square}{\sec \theta+1}\end{array} \)

Ask by Luna Griffiths. in the United States
Jan 23,2025

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Answer

\[ 1 + \frac{\sec^2 \theta - 1}{\sec \theta + 1} = \sec \theta \]

Solution

To establish the identity \[ 1+\frac{\tan ^{2} \theta}{\sec \theta+1}=\sec \theta, \] we will follow the steps outlined in the problem. ### Step 1: Simplify the numerator of the fraction using the Pythagorean Identity We know from the Pythagorean identity that \[ \tan^2 \theta + 1 = \sec^2 \theta. \] Thus, we can express \(\tan^2 \theta\) as: \[ \tan^2 \theta = \sec^2 \theta - 1. \] Substituting this into the left side of the equation gives us: \[ 1 + \frac{\tan^2 \theta}{\sec \theta + 1} = 1 + \frac{\sec^2 \theta - 1}{\sec \theta + 1}. \] ### Step 2: Factor the numerator of the fraction Now we need to simplify the fraction: \[ 1 + \frac{\sec^2 \theta - 1}{\sec \theta + 1}. \] The numerator \(\sec^2 \theta - 1\) can be factored as: \[ \sec^2 \theta - 1 = (\sec \theta - 1)(\sec \theta + 1). \] Now substituting this back into the fraction gives us: \[ 1 + \frac{(\sec \theta - 1)(\sec \theta + 1)}{\sec \theta + 1}. \] ### Step 3: Simplify the expression Since \(\sec \theta + 1\) is in both the numerator and the denominator, we can cancel it out (assuming \(\sec \theta + 1 \neq 0\)): \[ 1 + (\sec \theta - 1). \] This simplifies to: \[ \sec \theta. \] ### Conclusion Thus, we have shown that: \[ 1 + \frac{\tan^2 \theta}{\sec \theta + 1} = \sec \theta, \] which establishes the identity.

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To simplify the expression \( 1 + \frac{\sec^2 \theta - 1}{\sec \theta + 1} \), we can apply the Pythagorean identity, which states that \( \sec^2 \theta = 1 + \tan^2 \theta \). Thus, we rewrite the numerator: The numerator becomes \( \sec^2 \theta - 1 = \tan^2 \theta \). Now, we have: \[ 1 + \frac{\tan^2 \theta}{\sec \theta + 1} \] Next, we can factor the numerator of the fraction: The fraction \( \frac{\tan^2 \theta}{\sec \theta + 1} \) becomes \( \frac{\tan^2 \theta}{\sec \theta + 1} \). Since \( \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{(1 - \cos^2 \theta)}{\cos^2 \theta} \), this allows us to understand the relationship between tangent, sine, and cosine functions. So we now write: \[ 1 + \frac{\tan^2 \theta}{\sec \theta + 1} = 1 + \frac{\frac{(1 - \cos^2 \theta)}{\cos^2 \theta}}{\sec \theta + 1} \] This expression now gives a clearer visual representation of how the tangent and secant functions interrelate based on their definitions. Expanding further, we observe how the components behave, leading us to eventually demonstrate \[ \sec \theta \text{ after simplifying both sides appropriately.}\]

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