How do you evaluate $\sec {{110}^{\circ }}$?

Hint: The secant function is equal to the reciprocal of the cosine function. So we can write the given trigonometric expression as \[\dfrac{1}{\cos {{110}^{\circ }}}\]. Then, we have to use the trigonometric identity given by $\cos 3x=4{{\cos }^{3}}x-3\cos x$ and substitute $x={{110}^{\circ }}$ in the identity. On the LHS, we will get $\cos {{330}^{\circ }}$ which will be equal to $\dfrac{\sqrt{3}}{2}$ using the trigonometric identity $\cos \left( {{360}^{\circ }}-x \right)=\cos x$. On putting \[\cos {{110}^{\circ }}=x\], we will get a cubic equation in terms of \[x\] which can be solved using the graphing calculator to get the value of \[\cos {{110}^{\circ }}\]. Putting the value of \[\cos {{110}^{\circ }}\] in the expression \[\dfrac{1}{\cos {{110}^{\circ }}}\], we will get the final answer.

Complete step by step solution:
Let us consider the trigonometric expression given in the above question as
\[\Rightarrow E=\sec {{110}^{\circ }}\]
Now, we know that secant is the reciprocal of the cosine function, that is, \[\sec x=\dfrac{1}{\cos x}\]. Therefore, the above trigonometric expression can also be written as
\[\Rightarrow E=\dfrac{1}{\cos {{110}^{\circ }}}........\left( i \right)\]
So we have to evaluate the expression \[\cos {{110}^{\circ }}\]. For this we consider the trigonometric identity given by
$\Rightarrow \cos 3x=4{{\cos }^{3}}x-3\cos x$
Substituting $x={{110}^{\circ }}$ in the above identity, we get
\[\begin{align}
  & \Rightarrow \cos 3\left( {{110}^{\circ }} \right)=4{{\cos }^{3}}{{110}^{\circ }}-3\cos {{110}^{\circ }} \\
 & \Rightarrow \cos {{330}^{\circ }}=4{{\cos }^{3}}{{110}^{\circ }}-3\cos {{110}^{\circ }} \\
\end{align}\]
Writing \[{{330}^{\circ }}={{360}^{\circ }}-{{30}^{\circ }}\] in the LHS, we get
\[\Rightarrow \cos \left( {{360}^{\circ }}-{{30}^{\circ }} \right)=4{{\cos }^{3}}{{110}^{\circ }}-3\cos {{110}^{\circ }}\]
Now, we know that $\cos \left( {{360}^{\circ }}-x \right)=\cos x$. So the above expression becomes
\[\Rightarrow \cos {{30}^{\circ }}=4{{\cos }^{3}}{{110}^{\circ }}-3\cos {{110}^{\circ }}\]
We know that \[\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}\]. Putting this in the above equation, we get
\[\Rightarrow \dfrac{\sqrt{3}}{2}=4{{\cos }^{3}}{{110}^{\circ }}-3\cos {{110}^{\circ }}\]
Now let us substitute \[\cos {{110}^{\circ }}=x\] in the above equation to get
\[\Rightarrow \dfrac{\sqrt{3}}{2}=4{{x}^{3}}-3x\]
Subtracting \[\dfrac{\sqrt{3}}{2}\] from both the sides, we get
\[\begin{align}
  & \Rightarrow \dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{3}}{2}=4{{x}^{3}}-3x-\dfrac{\sqrt{3}}{2} \\
 & \Rightarrow 0=4{{x}^{3}}-3x-\dfrac{\sqrt{3}}{2} \\
 & \Rightarrow 4{{x}^{3}}-3x-\dfrac{\sqrt{3}}{2}=0 \\
\end{align}\]
Considering the graph of $y=4{{x}^{3}}-3x-\dfrac{\sqrt{3}}{2}$, we have

From the above graph, we get the roots as $x=-0.64,-0.34,0.98$. Since $x=\cos {{110}^{\circ }}$, the values $x=-0.34$ and $x=0.98$ are rejected. So we have
\[\begin{align}
  & \Rightarrow x=-0.64 \\
 & \Rightarrow \cos {{110}^{\circ }}=-0.64 \\
\end{align}\]
Putting this in (i) we get
$\begin{align}
  & \Rightarrow E=\dfrac{1}{-0.34} \\
 & \Rightarrow E=-2.94 \\
\end{align}$
Hence, the value of $\sec {{110}^{\circ }}$ is equal to $-2.94$.

Note: After solving the cubic equation obtained above, do not conclude it as your final answer. Since we had put \[\cos {{110}^{\circ }}=x\], the solution of the cubic equation will give the value of \[\cos {{110}^{\circ }}\]. But we have to find the value of \[\sec {{110}^{\circ }}\] which is equal to \[\dfrac{1}{\cos {{110}^{\circ }}}\].