Problem
Solve: \( \cot^{-1}x – \cot^{-1}(x+2) = \frac{\pi}{12}, \quad x > 0 \)
Solution
Let:
\[ A = \cot^{-1}x,\quad B = \cot^{-1}(x+2) \]
Step 1: Use identity
\[ \cot^{-1}x – \cot^{-1}y = \tan^{-1}\left(\frac{y – x}{1 + xy}\right) \]
So,
\[ \cot^{-1}x – \cot^{-1}(x+2) = \tan^{-1}\left(\frac{(x+2) – x}{1 + x(x+2)}\right) \]
\[ = \tan^{-1}\left(\frac{2}{x^2 + 2x + 1}\right) = \tan^{-1}\left(\frac{2}{(x+1)^2}\right) \]
Step 2: Compare with RHS
\[ \tan^{-1}\left(\frac{2}{(x+1)^2}\right) = \frac{\pi}{12} \]
Step 3: Convert to tan
\[ \frac{2}{(x+1)^2} = \tan\left(\frac{\pi}{12}\right) \]
\[ \tan\left(\frac{\pi}{12}\right) = 2 – \sqrt{3} \]
Step 4: Solve
\[ \frac{2}{(x+1)^2} = 2 – \sqrt{3} \]
\[ (x+1)^2 = \frac{2}{2 – \sqrt{3}} \]
Rationalize:
\[ = \frac{2(2 + \sqrt{3})}{1} = 4 + 2\sqrt{3} \]
\[ x+1 = \pm \sqrt{4 + 2\sqrt{3}} \]
Step 5: Use condition \(x>0\)
\[ x+1 = \sqrt{4 + 2\sqrt{3}} \]
\[ x = \sqrt{4 + 2\sqrt{3}} – 1 \]
Final Answer
\[ \boxed{\sqrt{4 + 2\sqrt{3}} – 1} \]
Explanation
Using cot⁻¹ subtraction identity and simplifying leads to a solvable equation.