If $\tan^{-1}(2n) - \tan^{-1}(n) = \tan^{-1}(2n)$, then find $n$.
Step-by-Step Solution
Key Concept: The constraint that a squared expression plus 1 must lie in $[-1, 1]$ forces it to equal 1, which uniquely determines the variable.
Starting with $z^2 - 2z + 2 = (z-1)^2 + 1 \geq 1$, we have $-1 \leq (z^2 - 2z + 2) \leq 1$ is only possible when $z^2 - 2z + 2 = 1$, which gives $z = 1$. We then verify: $a(1)^1 + \sin^{-1}(1) + \cos^{-1}(1) = 0$, so $a + \frac{\pi}{2} + 0 = 0$, yielding $a = -\frac{\pi}{2}$.
Correct Answer: C