Question

If \( A = 240^\circ \), then the value of \( \tan^2 A + \sec A \) is:

• A. \( \sqrt{3} \)
• B. 2
• C. 1
• D. 0

Hint:

In the third quadrant, \( \tan \) is positive, and \( \sec \) is negative. Use the known values for \( \tan 60^\circ \) and \( \cos 240^\circ \) to simplify the calculation.

Answer 1

The Correct Option is C

We are given that \( A = 240^\circ \). First, let's calculate \( \tan A \) and \( \sec A \) for \( A = 240^\circ \). Step 1: Calculate \( \tan A \) for \( A = 240^\circ \): - The tangent of \( 240^\circ \) is: \[ \tan 240^\circ = \tan (180^\circ + 60^\circ) = \tan 60^\circ = \sqrt{3} \] Since the angle is in the third quadrant, \( \tan 240^\circ \) is positive: \[ \tan 240^\circ = \sqrt{3} \] So, \[ \tan^2 240^\circ = (\sqrt{3})^2 = 3 \] Step 2: Calculate \( \sec A \) for \( A = 240^\circ \): - The secant is the reciprocal of cosine, and the cosine of \( 240^\circ \) is: \[ \cos 240^\circ = -\frac{1}{2} \] So, \[ \sec 240^\circ = \frac{1}{\cos 240^\circ} = \frac{1}{-\frac{1}{2}} = -2 \] Step 3: Now, calculate \( \tan^2 A + \sec A \): \[ \tan^2 240^\circ + \sec 240^\circ = 3 + (-2) = 1 \] Thus, the value of \( \tan^2 A + \sec A \) is 1.