## Precalculus (6th Edition) Blitzer

The solution is $3$.
Convert $\frac{\pi }{4}$ into degrees. \begin{align} & \frac{\pi }{4}=\frac{\pi }{4}\left( \frac{180\text{ }\!\!{}^\circ\!\!\text{ }}{\pi } \right) \\ & =\left( \frac{\pi }{\pi } \right)\left( \frac{180\text{ }\!\!{}^\circ\!\!\text{ }}{4} \right) \\ & =45\text{ }\!\!{}^\circ\!\!\text{ } \end{align} Convert $\frac{\pi }{6}$ into degrees. \begin{align} & \frac{\pi }{6}=\frac{\pi }{6}\left( \frac{180\text{ }\!\!{}^\circ\!\!\text{ }}{\pi } \right) \\ & =\left( \frac{\pi }{\pi } \right)\left( \frac{180\text{ }\!\!{}^\circ\!\!\text{ }}{6} \right) \\ & =30\text{ }\!\!{}^\circ\!\!\text{ } \end{align} The expressions for $\tan \theta$ and $\csc \theta$ are $\tan \theta =\frac{a}{b}$ …… (I) $\csc \theta =\frac{c}{a}$ …… (2) Here, $a$ is the length of the side opposite to $\theta$ , $b$ is the length of the side adjacent to $\theta$ and $c$ is the hypotenuse. In triangle 1, For $\theta =45\text{ }\!\!{}^\circ\!\!\text{ }$ , $a=1$ and $b=1$ Substitute $\frac{\pi }{4}$ for $\theta$ , $1$ for $a$ and $1$ for $b$ in equation (I). $\tan \frac{\pi }{4}=\frac{1}{1}$ …… (3) In triangle 2, For $\theta =30\text{ }\!\!{}^\circ\!\!\text{ }$ , $a=1$ and $c=2$ Substitute $\frac{\pi }{6}$ for $\theta$ , $1$ for $a$ and $2$ for $c$ in equation (2). $\csc \frac{\pi }{6}=\frac{2}{1}$ …… (4) Add equation (4) and equation (3). \begin{align} & \tan \frac{\pi }{4}+\csc \frac{\pi }{6}=\frac{1}{1}+\frac{2}{1} \\ & =3 \end{align} Therefore, the solution is $3$.