Question

\[ \frac{\tan^2 \theta + \cot^2 \theta + 2}{\sec \theta \csc \theta}, \ 0^\circ<\theta<90^\circ, \text{ is equal to:} \]

• A. \( \sin \theta \cos \theta \)
• B. \( \sec \theta \csc \theta \)
• C. \( \csc \theta \)
• D. \( \sec \theta \)

Hint:

For trigonometric simplifications, remember to use the basic trigonometric identities like \( \tan^2 \theta = \sec^2 \theta - 1 \), and \( \cot^2 \theta = \csc^2 \theta - 1 \).

Answer 1

The Correct Option is B

We are given the expression: \[ \frac{\tan^2 \theta + \cot^2 \theta + 2}{\sec \theta \csc \theta}. \] By using standard trigonometric identities and simplification: \[ \tan^2 \theta + \cot^2 \theta = \sec^2 \theta - 1 + \csc^2 \theta - 1. \] Now simplify the whole expression, and it leads to: \[ \sec \theta \csc \theta. \]