A solution of the equation $\cot^{-1} 2 = \cot^{-1} x + \cot^{-1}(10-x)$ where $1 < x < 9$ is:
Step-by-Step Solution
Key Concept: The addition formula for inverse cotangent converts the sum into a single inverse cotangent, which can then be equated to the right side.
We use the identity $\cot^{-1} a + \cot^{-1} b = \cot^{-1}\left(\frac{ab-1}{a+b}\right)$ when $ab > 1$. Applying this: $\cot^{-1} x + \cot^{-1}(10-x) = \cot^{-1}\left(\frac{x(10-x)-1}{x+(10-x)}\right) = \cot^{-1}\left(\frac{10x-x^2-1}{10}\right)$. Setting this equal to $\cot^{-1} 2$: $\frac{10x-x^2-1}{10} = 2$, which gives $10x - x^2 - 1 = 20$, or $x^2 - 10x + 21 = 0$. Factoring: $(x-7)(x-3) = 0$, so $x = 7$ or $x = 3$. Both solutions satisfy $1 < x < 9$ and the condition $x(10-x) > 1$.
Correct Answer: 1,2