Question

If \( \frac{1-\tan\theta}{1+\tan\theta} = \frac{1}{\sqrt{3}}, \) where \( \theta \in \left(0,\frac{\pi}{2}\right) \), then \( \theta = \)

• A. \( \dfrac{\pi}{12} \)
• B. \( \dfrac{\pi}{4} \)
• C. \( \dfrac{\pi}{6} \)
• D. \( \dfrac{\pi}{3} \)

Hint:

Expressions of the form \( \frac{1-\tan\theta}{1+\tan\theta} \) can be simplified using the identity \( \tan\left(\frac{\pi}{4}-\theta\right) \).

Answer 1

The Correct Option is A

Step 1: Use the tangent subtraction identity.
We know that \[ \frac{1-\tan\theta}{1+\tan\theta} = \tan\left(\frac{\pi}{4}-\theta\right) \] Step 2: Rewrite the given equation.
\[ \tan\left(\frac{\pi}{4}-\theta\right) = \frac{1}{\sqrt{3}} \] Step 3: Use the standard value of inverse tangent.
\[ \frac{\pi}{4}-\theta = \frac{\pi}{6} \] Step 4: Solve for \( \theta \).
\[ \theta = \frac{\pi}{4}-\frac{\pi}{6} = \frac{\pi}{12} \] Step 5: Conclusion.
Since \( \theta \in \left(0,\frac{\pi}{2}\right) \), the required value is \[ \theta = \frac{\pi}{12} \]