Question

If \( \sec x = \frac{5}{4} \), then \( \frac{\tan x}{1 + \tan^2 x} \) is equal to:

• A. \( \frac{1}{25} \)
• B. \( \frac{9}{25} \)
• C. \( \frac{3}{4} \)
• D. \( \frac{12}{25} \)

Hint:

Remember that \( \sec^2 x - 1 = \tan^2 x \), and always simplify the trigonometric identities before solving.

Answer 1

The Correct Option is B

We are given that \( \sec x = \frac{5}{4} \), which implies that \( \cos x = \frac{4}{5} \). Using the identity: \[ \tan^2 x = \sec^2 x - 1 \] Substituting the value of \( \sec x \), we get: \[ \tan^2 x = \left(\frac{5}{4}\right)^2 - 1 = \frac{25}{16} - 1 = \frac{9}{16}. \] Thus, \( \tan x = \frac{3}{4} \). Now, using the identity: \[ \frac{\tan x}{1 + \tan^2 x} = \frac{\frac{3}{4}}{1 + \frac{9}{16}} = \frac{\frac{3}{4}}{\frac{25}{16}} = \frac{3}{4} \times \frac{16}{25} = \frac{9}{25}. \] Therefore, the correct answer is \( \frac{9}{25} \).