Problem
Solve: \( \tan^{-1}\left(\frac{1-x}{1+x}\right) – \frac{1}{2}\tan^{-1}(x) = 0,\quad x > 0 \)
Solution
Step 1: Rearrange
\[ \tan^{-1}\left(\frac{1-x}{1+x}\right) = \frac{1}{2}\tan^{-1}(x) \]
Step 2: Use identity
\[ \tan^{-1}\left(\frac{1-x}{1+x}\right) = \frac{\pi}{4} – \tan^{-1}(x) \quad (x>0) \]
Step 3: Substitute
\[ \frac{\pi}{4} – \tan^{-1}(x) = \frac{1}{2}\tan^{-1}(x) \]
Step 4: Solve
\[ \frac{\pi}{4} = \frac{3}{2}\tan^{-1}(x) \]
\[ \tan^{-1}(x) = \frac{\pi}{6} \]
Step 5: Find x
\[ x = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \]
Step 6: Check condition
Since \( x > 0 \), the solution is valid.
Final Answer
\[ \boxed{\frac{1}{\sqrt{3}}} \]
Explanation
Using identity tan⁻¹((1−x)/(1+x)) = π/4 − tan⁻¹x (for x > 0), we reduce to a simple linear equation.