## Precalculus (6th Edition) Blitzer

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RECALL: (1)The reference angle of a given angle is equal to the smallest acute angle that the terminal side makes with the x-axis. (2) Based on the location of the terminal side of an angle $\theta$, the reference angle can be found using the formula: (i) Quadrant I: $\theta$ (ii) Quadrant II: $180^0-\theta$ (iii) Quadrant III: $\theta-180^o$ (iv) Quadrant IV: $360^o-\theta$ The given angle is coterminal with $405^o-360^o=45^o$. This angle is in Quadrant I so its reference angle is itself. Thus, $\tan{405^o} = \tan{45^o}$ $45^o$ is a special angle whose tangent value is $1$. Thus, $\tan{405^o}=\tan{45^o}=1$