Problem
Solve: \( \tan^{-1}x + 2\cot^{-1}x = \frac{2\pi}{3} \)
Solution
Step 1: Use identity
\[ \cot^{-1}x = \frac{\pi}{2} – \tan^{-1}x \]
Step 2: Substitute
\[ \tan^{-1}x + 2\left(\frac{\pi}{2} – \tan^{-1}x\right) = \frac{2\pi}{3} \]
\[ \tan^{-1}x + \pi – 2\tan^{-1}x = \frac{2\pi}{3} \]
\[ \pi – \tan^{-1}x = \frac{2\pi}{3} \]
Step 3: Solve
\[ \tan^{-1}x = \frac{\pi}{3} \]
Step 4: Find x
\[ x = \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \]
Step 5: Check principal range
Since \( \tan^{-1}x \in (-\frac{\pi}{2}, \frac{\pi}{2}) \), the solution is valid.
Final Answer
\[ \boxed{\sqrt{3}} \]
Explanation
Convert cot⁻¹x into tan⁻¹x and solve the linear equation.