b) \( \tan 405^{\circ}-\tan (\arctan 1)= \)

To solve \( \tan 405^{\circ} - \tan (\arctan 1) \), we first simplify \( \tan 405^{\circ} \). Since the tangent function has a periodicity of \( 180^{\circ} \), we can reduce \( 405^{\circ} \) by subtracting \( 360^{\circ} \): \[ 405^{\circ} - 360^{\circ} = 45^{\circ} \] Thus, \( \tan 405^{\circ} = \tan 45^{\circ} = 1 \). Next, recall that \( \tan (\arctan 1) = 1 \), since the tangent function and the arctangent function are inverses. Putting it all together: \[ \tan 405^{\circ} - \tan (\arctan 1) = 1 - 1 = 0 \] So, the final answer is: \[ \boxed{0} \]