Question

If \( \tan^{-1}\!\left(\dfrac{1-x}{1+x}\right) - \dfrac{1{2}\tan^{-1}x = 0 \), for \( x>0 \), then the value of \( x \) is}

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

Hint:

Inverse trigonometric equations can often be simplified using standard angle identities before solving.

Answer 1

The Correct Option is C

Step 1: Rewrite the given equation.
\[ \tan^{-1}\!\left(\frac{1-x}{1+x}\right) = \frac{1}{2}\tan^{-1}x \]

Step 2: Use the identity for inverse tangent.
For \( x>0 \), \[ \tan^{-1}\!\left(\frac{1-x}{1+x}\right) = \frac{\pi}{4} - \tan^{-1}x \]

Step 3: Substitute and simplify.
\[ \frac{\pi}{4} - \tan^{-1}x = \frac{1}{2}\tan^{-1}x \] \[ \frac{\pi}{4} = \frac{3}{2}\tan^{-1}x \] \[ \tan^{-1}x = \frac{\pi}{6} \]

Step 4: Find the value of \( x \).
\[ x = \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}} \]

Step 5: Conclusion.
The required value of \( x \) is \[ \boxed{\dfrac{1}{\sqrt{3}}} \]