Question

The principal value of $\tan^{-1}\left(\cot\left(\frac{43\pi}{4}\right)\right)$ is

• A. $\frac{-3\pi}{4}$
• B. $\frac{3\pi}{4}$
• C. $\frac{-\pi}{4}$
• D. $\frac{\pi}{4}$

Answer 1

Answer: (C)

$\frac{-\pi}{4}$

Answer 2

1. Write

   $$\cot\left(\frac{43\pi}{4}\right)=\tan\left(\frac{\pi}{2}-\frac{43\pi}{4}\right)=\tan\left(-\frac{41\pi}{4}\right).$$

2. Since the principal value of $\tan^{-1}$ lies in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, we find an equivalent angle:

   $$-\frac{41\pi}{4} + 10\pi = -\frac{41\pi}{4} + \frac{40\pi}{4} = -\frac{\pi}{4}.$$

3. Thus,

   $$\tan^{-1}\left(\cot\left(\frac{43\pi}{4}\right)\right) = -\frac{\pi}{4}.$$

Explanation (Minimal): Express $\cot\left(\frac{43\pi}{4}\right)$ as $\tan\left(-\frac{41\pi}{4}\right)$. Adjust $-\frac{41\pi}{4}$ by adding $10\pi$ (due to the $\pi$ period of $\tan$) to yield $-\frac{\pi}{4}$ which lies in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.