Establish the identity. \( \begin{array}{l}1+\frac{\tan ^{2} \theta}{\sec \theta+1}=\sec \theta \\ \text { Simplify the numerator of the fraction from the left side by applying the appropriate Pythagorean Identity. } \\ 1+\frac{\sec ^{2} \theta-1}{\sec \theta+1} \text { (Do not factor.) } \\ \text { Factor the numerator of the fraction from the previous step. } \\ 1+\frac{\square}{\sec \theta+1}\end{array} \)
Solución de tutoría real
Responder
Solución
Revisado y aprobado por el equipo de tutoría de UpStudy
The Deep Dive
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 \[ \sec \theta \text{ after simplifying both sides appropriately.}\]