## Precalculus (6th Edition) Blitzer

The value of $6\tan \frac{\pi }{4}+\sin \frac{\pi }{3}\sec \frac{\pi }{6}$ is $7$
Consider the given expression $6\tan \frac{\pi }{4}+\sin \frac{\pi }{3}\sec \frac{\pi }{6}$. Consider the provided right angled triangle. To find $\tan \left( \frac{\pi }{4} \right)$ apply the definition of $\tan \theta$ to the provided triangle, Therefore, $\tan \theta =\frac{\text{Length of side opposite of }\theta }{\text{Length of side adjacent of }\theta }$. But, here, $\theta =\frac{\pi }{4}$ and $\tan \left( \frac{\pi }{4} \right)=\tan 45{}^\circ$. Therefore, \begin{align} & \tan \left( \frac{\pi }{4} \right)=\frac{\text{Length of side opposite of 45}{}^\circ }{\text{Length of side adjacent of 45}{}^\circ } \\ & =\frac{1}{1} \\ & =1 \end{align} Now, to find $\sin \left( \frac{\pi }{3} \right)$ apply the definition of $\sin \theta$ to the above triangle, $\sin \theta =\frac{\text{Length of side opposite of }\theta }{\text{hypotenuse}}$. But, here, $\theta =\frac{\pi }{3}$ , and $\sin \left( \frac{\pi }{3} \right)=\sin 60{}^\circ$. Therefore, \begin{align} & \sin \left( \frac{\pi }{3} \right)=\frac{\text{Length of side opposite of 60}{}^\circ }{\text{Length of hypotenuse}} \\ & =\frac{\sqrt{3}}{2} \end{align} To find $\cos \left( \frac{\pi }{6} \right)$ apply the definition of $\cos \theta$ to the provided triangle, $\cos \theta =\frac{\text{Length of side adjacent of }\theta }{\text{hypotenuse}}$. But, here, $\theta =\frac{\pi }{6}$ , and $\cos \left( \frac{\pi }{6} \right)=\cos 30{}^\circ$. Therefore, \begin{align} & \cos \left( \frac{\pi }{6} \right)=\frac{\text{Length of side adjacent of 30}{}^\circ }{\text{Length of hypotenuse}} \\ & =\frac{\sqrt{3}}{2} \end{align} Now, apply the formula for the inverse trigonometric function as, $\sec \theta =\frac{1}{\cos \theta }$. \begin{align} & \sec \left( \frac{\pi }{6} \right)=\frac{1}{\cos \left( \frac{\pi }{6} \right)} \\ & =\frac{1}{\left( \frac{\sqrt{3}}{2} \right)} \\ & =\frac{2}{\sqrt{3}} \end{align} Rationalize the denominator as, \begin{align} & \sec \left( \frac{\pi }{6} \right)=\frac{2}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}} \\ & =\frac{2\sqrt{3}}{3} \end{align} Now substitute $\tan \left( \frac{\pi }{4} \right)=1,\text{ sin}\left( \frac{\pi }{3} \right)=\frac{\sqrt{3}}{2}\text{ and sec}\left( \frac{\pi }{6} \right)=\frac{2\sqrt{3}}{3}$ in the provided expression, $6\tan \frac{\pi }{4}+\sin \frac{\pi }{3}\sec \frac{\pi }{6}$. Therefore, \begin{align} & 6\tan \frac{\pi }{4}+\sin \frac{\pi }{3}\sec \frac{\pi }{6}=6\left( 1 \right)+\left( \frac{\sqrt{3}}{2} \right)\left( \frac{2\sqrt{3}}{3} \right) \\ & =6+\frac{2\sqrt{9}}{6} \\ & =6+\frac{2\left( 3 \right)}{6} \\ & =6+\frac{6}{6} \end{align} Further simplify, \begin{align} & 6\tan \frac{\pi }{4}+\sin \frac{\pi }{3}\sec \frac{\pi }{6}=6+1 \\ & =7 \end{align} Hence the value of $6\tan \frac{\pi }{4}+\sin \frac{\pi }{3}\sec \frac{\pi }{6}$ is $7$.