Question

If \( \frac{\sec^2 15^\circ - 1}{\sec^2 15^\circ} \) equals:

• A. \( \frac{2 - \sqrt{3}}{4} \)
• B. \( \frac{2 + \sqrt{3}}{4} \)
• C. \( \frac{2 - \sqrt{3}}{2} \)
• D. \( \frac{2 + \sqrt{3}}{2} \)
• E. \( \frac{1}{4} \)

Hint:

Utilize trigonometric identities to simplify expressions involving angles and their functions.

Answer 1

The Correct Option is A

The given expression can be simplified using the trigonometric identity \( \sec^2 \theta = 1 + \tan^2 \theta \): \[ \frac{\sec^2 15^\circ - 1}{\sec^2 15^\circ} = \frac{\tan^2 15^\circ}{\sec^2 15^\circ} \] Using the identity \( \sec^2 \theta = \frac{1}{\cos^2 \theta} \), the expression simplifies to: \[ \frac{\tan^2 15^\circ}{\frac{1}{\cos^2 15^\circ}} = \tan^2 15^\circ \cdot \cos^2 15^\circ = \sin^2 15^\circ \] The value of \( \sin 15^\circ \) can be calculated using the formula \( \sin 15^\circ = \sin(45^\circ - 30^\circ) \) and the sine addition formula: \[ \sin 15^\circ = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ \] \[ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} - \sqrt{2}}{4} \] Squaring this to find \( \sin^2 15^\circ \): \[ \sin^2 15^\circ = \left(\frac{\sqrt{6} - \sqrt{2}}{4}\right)^2 = \frac{6 - 2\sqrt{12} + 2}{16} = \frac{8 - 4\sqrt{3}}{16} = \frac{2 - \sqrt{3}}{4} \] Thus, the answer is: \[ \frac{2 - \sqrt{3}}{4} \]