Question

If \[ \frac{1}{1 - \tan x} = \frac{3 + \sqrt{3}}{2}, \quad 0 \leq x \leq \frac{\pi}{2}, \] then the value of \( x \) is equal to:

• A. \( \frac{\pi}{3} \)
• B. \( \frac{\pi}{5} \)
• C. \( \frac{\pi}{6} \)
• D. \( \frac{\pi}{8} \)
• E. \( \frac{\pi}{12} \)

Hint:

When given an equation involving \( \tan x \), rationalize the denominator or use known trigonometric values to solve for \( x \).

Answer 1

The Correct Option is C

We are given that: \[ \frac{1}{1 - \tan x} = \frac{3 + \sqrt{3}}{2} \] Step 1: Simplifying the equation
Rearrange the equation to express \( 1 - \tan x \) as: \[ 1 - \tan x = \frac{2}{3 + \sqrt{3}} \] Now, rationalize the denominator by multiplying the numerator and denominator by \( 3 - \sqrt{3} \): \[ 1 - \tan x = \frac{2}{3 + \sqrt{3}} \times \frac{3 - \sqrt{3}}{3 - \sqrt{3}} = \frac{2(3 - \sqrt{3})}{(3 + \sqrt{3})(3 - \sqrt{3})} \] Simplifying the denominator using the difference of squares formula: \[ (3 + \sqrt{3})(3 - \sqrt{3}) = 9 - 3 = 6 \] Thus: \[ 1 - \tan x = \frac{2(3 - \sqrt{3})}{6} = \frac{3 - \sqrt{3}}{3} \] Step 2: Solving for \( \tan x \) Now, solve for \( \tan x \): \[ \tan x = 1 - \frac{3 - \sqrt{3}}{3} = \frac{3}{3} - \frac{3 - \sqrt{3}}{3} = \frac{\sqrt{3}}{3} \] Thus: \[ \tan x = \frac{1}{\sqrt{3}} \] Step 3: Finding the value of \( x \) We know that: \[ \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} \] Thus: \[ x = \frac{\pi}{6} \] Thus, the correct answer is option (C), \( x = \frac{\pi}{6} \).