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**Hint**: The given question deals with basic simplification of trigonometric functions by using some of the simple trigonometric formulae. All the trigonometric formulae, rules and identities must be remembered by heart before solving such a complex question. Basic algebraic rules and simplification techniques are to be kept in mind while doing simplification in the given problem.

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__Complete step-by-step answer__In the given problem, we have to find the value of the trigonometric expression $\cos {10^ \circ } + \cos {110^ \circ } + \cos {130^ \circ }$.

We can simplify the trigonometric expression given to us by using the trigonometric formula involving the sum of two cosines of different angles. First, we have to group the cosine functions on which we will apply the identity. So, we get,

$ \Rightarrow \left( {\cos {{10}^ \circ } + \cos {{110}^ \circ }} \right) + \cos {130^ \circ }$

Now, using the trigonometric identity $\cos C + \cos D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$ , we get,

$ \Rightarrow 2\cos \left( {\dfrac{{{{10}^ \circ } + {{110}^ \circ }}}{2}} \right)\cos \left( {\dfrac{{{{10}^ \circ } - {{110}^ \circ }}}{2}} \right) + \cos {130^ \circ }$

Simplifying the expression, we get,

$ \Rightarrow 2\cos \left( {\dfrac{{{{120}^ \circ }}}{2}} \right)\cos \left( { - \dfrac{{{{100}^ \circ }}}{2}} \right) + \cos {130^ \circ }$

$ \Rightarrow 2\cos \left( {{{60}^ \circ }} \right)\cos \left( { - {{50}^ \circ }} \right) + \cos {130^ \circ }$

Now, we know that $\cos \left( { - x} \right) = \cos x$. So, using this in the expression, we get,

$ \Rightarrow 2\cos {60^ \circ }\cos {50^ \circ } + \cos {130^ \circ }$

We also know that the value of $\cos {60^ \circ }$ as $\left( {\dfrac{1}{2}} \right)$. So, we get,

$ \Rightarrow 2 \times \dfrac{1}{2}\cos {50^ \circ } + \cos {130^ \circ }$

Canceling the common factors in numerator and denominator and using the trigonometric formula $\cos \left( {{{180}^ \circ } - x} \right) = - \cos x$, we get,

$ \Rightarrow \cos {50^ \circ } + \cos {130^ \circ }$

We can write \[{130^ \circ }\] as \[{180^ \circ } - {50^ \circ }\] . So, we get,

$ \Rightarrow \cos {50^ \circ } + \cos \left( {{{180}^ \circ } - {{50}^ \circ }} \right)$

\[ \Rightarrow \cos {50^ \circ } - \cos {50^ \circ }\]

So, cancelling the like terms with opposite signs, we get,

\[ \Rightarrow 0\]

Hence, the value of $\cos {10^ \circ } + \cos {110^ \circ } + \cos {130^ \circ }$ is zero.

**Note**: Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such types of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations. However, questions involving this type of simplification of trigonometric ratios may also have multiple interconvertible answers.

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