Find \( \arctan \left(\tan 3^{*}\right. \) pi / 4) Find \( \arctan \left(\tan 2^{*}\right. \) pi / 3)

To solve for \( \arctan \left( \tan 3 \frac{\pi}{4} \right) \), first recall that \( \tan \left( \frac{5\pi}{4} \right) = \tan \left( 3 \frac{\pi}{4} \right) \) since \( \tan \) has a periodicity of \( \pi \). Therefore, we simplify \( 3 \frac{\pi}{4} \) to \( \frac{5\pi}{4} \). Now, since \( \tan \) is negative in the second quadrant (where \( 3 \frac{\pi}{4} \) lies), we have: \[ \tan \left( 3 \frac{\pi}{4} \right) = -1 \] Thus, \( \arctan \left( -1 \right) = -\frac{\pi}{4} \) (keeping the principal value). Therefore, \[ \arctan \left( \tan 3 \frac{\pi}{4} \right) = -\frac{\pi}{4} \] Now for \( \arctan \left( \tan 2 \frac{\pi}{3} \right) \). We can rewrite \( 2 \frac{\pi}{3} \) as \( \frac{2\pi}{3} = \frac{4\pi}{3} - \pi \) which corresponds to an angle in the second quadrant: \[ \tan \left( 2 \frac{\pi}{3} \right) = -\sqrt{3} \] Since \( \arctan \left( -\sqrt{3} \right) \) is the angle that gives us that tangent in the principal range of \( (-\frac{\pi}{2}, \frac{\pi}{2}) \), it yields: \[ \arctan \left( \tan 2 \frac{\pi}{3} \right) = -\frac{\pi}{3} \] To summarize the results: 1. \( \arctan \left( \tan 3 \frac{\pi}{4} \right) = -\frac{\pi}{4} \) 2. \( \arctan \left( \tan 2 \frac{\pi}{3} \right) = -\frac{\pi}{3} \)