Certainly! Let's simplify the given expression step by step.
The given expression is:
[tex]\[ 5 \frac{\sqrt{1 + \tan^2 \theta}}{\sqrt{1 - \sin^2 \theta}} \][/tex]
First, recall the fundamental trigonometric identities:
1. Pythagorean Identity for tangent and secant:
[tex]\[ 1 + \tan^2 \theta = \sec^2 \theta \][/tex]
2. Another Pythagorean Identity for sine:
[tex]\[ 1 - \sin^2 \theta = \cos^2 \theta \][/tex]
Using these identities, we can rewrite the expression:
[tex]\[ 5 \frac{\sqrt{1 + \tan^2 \theta}}{\sqrt{1 - \sin^2 \theta}} = 5 \frac{\sqrt{\sec^2 \theta}}{\sqrt{\cos^2 \theta}} \][/tex]
Since [tex]\( \sqrt{\sec^2 \theta} = \sec \theta \)[/tex] and [tex]\( \sqrt{\cos^2 \theta} = \cos \theta \)[/tex], the expression simplifies to:
[tex]\[ 5 \frac{\sec \theta}{\cos \theta} \][/tex]
Next, recall that [tex]\( \sec \theta = \frac{1}{\cos \theta} \)[/tex]:
[tex]\[ 5 \frac{\frac{1}{\cos \theta}}{\cos \theta} = 5 \frac{1}{\cos \theta \cdot \cos \theta} = 5 \frac{1}{\cos^2 \theta} \][/tex]
Since [tex]\( \frac{1}{\cos^2 \theta} = \sec^2 \theta \)[/tex], we have:
[tex]\[ 5 \sec^2 \theta \][/tex]
Thus, the simplified form of the given expression is:
[tex]\[ 5 \sec^2 \theta \][/tex]
