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