Problem
Evaluate: \( \sin\left(\tan^{-1}x + \tan^{-1}\frac{1}{x}\right), \quad x > 0 \)
Solution
Let:
\[ A = \tan^{-1}x,\quad B = \tan^{-1}\frac{1}{x} \]
Step 1: Use identity
\[ \tan^{-1}x + \tan^{-1}\frac{1}{x} = \begin{cases} \frac{\pi}{2}, & x > 0 \\ -\frac{\pi}{2}, & x < 0 \end{cases} \]
Since \( x > 0 \):
\[ A + B = \frac{\pi}{2} \]
Step 2: Evaluate sine
\[ \sin\left(A + B\right) = \sin\left(\frac{\pi}{2}\right) \]
\[ = 1 \]
Final Answer
\[ \boxed{1} \]
Explanation
For x > 0, the sum equals π/2 and sine of π/2 is 1.