Answer
The value of $\cos \frac{\pi }{3}\sec \frac{\pi }{3}-\cot \frac{\pi }{3}$ is $\frac{3-\sqrt{3}}{3}$.
Work Step by Step
Consider the given expression $\cos \frac{\pi }{3}\sec \frac{\pi }{3}-\cot \frac{\pi }{3}$.
Consider the provided right angled triangle.
To find $\cos \left( \frac{\pi }{3} \right)$ apply the definition of $\cos \theta $ to the provided triangle,
$\cos \theta =\frac{\text{Length of side adjacent of }\theta }{\text{hypotenuse}}$.
But, here, $\theta =\frac{\pi }{3}$ , and $\cos \left( \frac{\pi }{3} \right)=\cos 60{}^\circ $.
Therefore,
$\begin{align}
& \cos \left( \frac{\pi }{3} \right)=\frac{\text{Length of side adjacent of 60}{}^\circ }{\text{Length of hypotenuse}} \\
& =\frac{1}{2}
\end{align}$
Now, apply the formula for the inverse trigonometric function as, $\sec \theta =\frac{1}{\cos \theta }$.
$\begin{align}
& \sec \left( \frac{\pi }{3} \right)=\frac{1}{\cos \left( \frac{\pi }{3} \right)} \\
& =\frac{1}{\left( \frac{1}{2} \right)} \\
& =2
\end{align}$
Now, apply the definition of $\tan \theta $ to the provided triangle,
$\tan \theta =\frac{\text{Length of side opposite of }\theta }{\text{Length of side adjacent of }\theta }$.
Therefore,
$\begin{align}
& \tan \left( \frac{\pi }{3} \right)=\frac{\text{Length of side opposite of 60}{}^\circ }{\text{Length of side adjacent of 60}{}^\circ } \\
& =\frac{\sqrt{3}}{1} \\
& =\sqrt{3}
\end{align}$
But,
$\begin{align}
& \cot \left( \frac{\pi }{3} \right)=\frac{1}{\tan \left( \frac{\pi }{3} \right)} \\
& =\frac{1}{\sqrt{3}}
\end{align}$
Rationalize the denominator as,
$\begin{align}
& \cot \left( \frac{\pi }{3} \right)=\frac{1}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}} \\
& =\frac{\sqrt{3}}{3}
\end{align}$
Now substitute $\cos \left( \frac{\pi }{3} \right)=\frac{1}{2},\text{ sec}\left( \frac{\pi }{3} \right)=2\ \text{and }\cot \left( \frac{\pi }{3} \right)=\frac{\sqrt{3}}{3}$ in the provided expression, $\cos \frac{\pi }{3}\sec \frac{\pi }{3}-\cot \frac{\pi }{3}$.
Therefore,
$\begin{align}
& \cos \frac{\pi }{3}\sec \frac{\pi }{3}-\cot \frac{\pi }{3}=\frac{1}{2}.\left( 2 \right)-\frac{\sqrt{3}}{3} \\
& =1-\frac{\sqrt{3}}{3} \\
& =\frac{3-\sqrt{3}}{3}
\end{align}$
Hence the value of $\cos \frac{\pi }{3}\sec \frac{\pi }{3}-\cot \frac{\pi }{3}$ is $\frac{3-\sqrt{3}}{3}$.
