Answer
The value of $6\tan \frac{\pi }{4}+\sin \frac{\pi }{3}\sec \frac{\pi }{6}$ is $7$
Work Step by Step
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 $7$.
