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**Hint:**By the help of trigonometry table we can simply put the values of the trigonometric functions in the equation

After that we will try to simplify it to get the answer.

Trigonometric functions (or circular functions) are defined as the functions of an angle of a right-angled triangle.

**Complete step-by-step answer:**By the help of the trigonometry table, we have the values of all the functions used in the equation as follows:

\[\operatorname{Sin} {0^ \circ } = 1\] , \[\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}\],

\[\operatorname{Sin} {30^ \circ } = \dfrac{1}{2},{\text{ }}\tan {45^ \circ } = 1\]

\[\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }},{\text{ }}\cot {60^ \circ } = \dfrac{1}{{\sqrt 3 }}\]

\[

\\

\sec {60^ \circ } = 2,{\text{ }}\cos ec{90^ \circ } = 1 \\

\]

Now, Let us take the Left Hand side of the given equation:

L H S \[ = \dfrac{{(\operatorname{Sin} {0^ \circ } + \operatorname{Cos} {{30}^ \circ })(\operatorname{Sin} {{30}^ \circ } + \tan {{45}^ \circ })}}{{(\tan {{30}^ \circ } + \cot {{60}^ \circ })(\sec {{60}^ \circ } - \cos ec{{90}^ \circ })}}\]

On substituting all the above values of trigonometric functions in the L H S of the equation, we get:

\[

= \dfrac{{\left( {0 + \dfrac{{\sqrt 3 }}{2}} \right)\left( {\dfrac{1}{2} + 1} \right)}}{{\left( {\dfrac{1}{{\sqrt 3 }} + \dfrac{1}{{\sqrt 3 }}} \right)\left( {2 - 1} \right)}} \\

\\

\]

\[ = \dfrac{{\dfrac{{\sqrt 3 }}{2}\left( {\dfrac{3}{2}} \right)}}{{\dfrac{2}{{\sqrt 3 }}}}\]

Simplifying it further, we get:

\[

= \dfrac{{3\sqrt 3 }}{4} \times \dfrac{{\sqrt 3 }}{2} \\

= \dfrac{9}{8} \\

\]

= RHS

Which is equal to the right hand side of the equation.

**The given statement is proved.**

**Note:**Trigonometry is a branch of mathematics that studies the relationships between the side lengths and the angles of triangles.

There are 6 basic types of trigonometric functions which are:

sin function

cos function

tan function

cot function

cosec function

sec function.

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