Evaluate the expression (i) \[\sin {60^ \circ }\cos {30^ \circ } + \sin {30^ \circ }\cos {60^ \circ }\]
(ii) $\dfrac{{5{{\cos }^2}{{60}^ \circ } + 4{{\sec }^2}{{30}^ \circ } - {{\tan }^2}{{45}^ \circ }}}{{{{\sin }^2}{{30}^ \circ } + {{\cos }^2}{{30}^ \circ }}}$
(A) (i) $7$
(ii) $\dfrac{5}{{32}}$
(B) (i) $2$
(ii) $\dfrac{{33}}{8}$
(C) (i) $1$
(ii) $\dfrac{{67}}{{12}}$
(D) (i) $5$
(ii) $\dfrac{6}{{47}}$

Hint: There are two methods to solve these problems the first one involves the use of trigonometric identity.
$\sin (A + B) = \sin A\cos B + \cos A\sin B$ , the given equation is of the form of the right hand side of his trigonometric identity. The second method to solving is explained at the later end.

Complete step-by-step solution:
(i) Given trigonometric equation is
\[\sin {60^ \circ }\cos {30^ \circ } + \sin {30^ \circ }\cos {60^ \circ }\]
As we know that $\sin (A + B) = \sin A\cos B + \cos A\sin B$ so use this property in above equation we have,
Where $A = {60^ \circ }$ , $B = {30^ \circ }$ .
$ \Rightarrow \sin {60^ \circ }\cos {30^ \circ } + \sin {30^ \circ }\cos {60^ \circ } = \sin ({60^ \circ } + {30^ \circ })$
$ \Rightarrow \sin {60^ \circ }\cos {30^ \circ } + \sin {30^ \circ }\cos {60^ \circ } = \sin ({90^ \circ })$
As we know the value of $\sin {90^ \circ } = 1$ , put this in above equation and we get
$ \Rightarrow \sin {60^ \circ }\cos {30^ \circ } + \sin {30^ \circ }\cos {60^ \circ } = 1$
So the required value of the given trigonometric equation (i) is $1$ .
(ii) The given equation is $\dfrac{{5{{\cos }^2}{{60}^ \circ } + 4{{\sec }^2}{{30}^ \circ } - {{\tan }^2}{{45}^ \circ }}}{{{{\sin }^2}{{30}^ \circ } + {{\cos }^2}{{30}^ \circ }}}$
As we know that ${\cos ^2}\theta = 1 - {\sin ^2}\theta $ , use this in above equation and we get
\[ = \dfrac{{5(1 - {{\sin }^2}{{60}^ \circ }) + 4{{\sec }^2}{{30}^ \circ } - {{\tan }^2}{{45}^ \circ }}}{{{{\sin }^2}{{30}^ \circ } + {{\cos }^2}{{30}^ \circ }}}\]
Also we know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$ , use this in above equation and we get
$ = \dfrac{{5(1 - {{\sin }^2}{{60}^ \circ }) + 4{{\sec }^2}{{30}^ \circ } - {{\tan }^2}{{45}^ \circ }}}{1}$
$ = 5(1 - {\sin ^2}{60^ \circ }) + 4{\sec ^2}{30^ \circ } - {\tan ^2}{45^ \circ }$
$ = 5 - 5{\sin ^2}{60^ \circ } + 4{\sec ^2}{30^ \circ } - {\tan ^2}{45^ \circ }$
\[ = 5 - 5{\sin ^2}{60^ \circ } + 4 \times \dfrac{1}{{{{\cos }^2}{{30}^ \circ }}} - {\tan ^2}{45^ \circ }\]
We know that the value of $\sin {60^ \circ } = \cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$ and $\tan {45^ \circ } = 1$ , we get
$ = 5 - 5 \times {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} + 4 \times \dfrac{1}{{{{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}^2}}} - {\left( 1 \right)^2}$
We know that ${\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} = \dfrac{3}{4}$ and $\dfrac{1}{{{{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}^2}}} = \dfrac{1}{{\dfrac{3}{4}}} = \dfrac{4}{3}$ , use this and we get
$ = 5 - 5 \times \dfrac{3}{4} + 4 \times \dfrac{4}{3} - 1$
Simplifying and we get
$ = 5 - \dfrac{{15}}{4} + \dfrac{{16}}{3} - 1$
$ = \dfrac{{60 - 45 + 64 - 12}}{{12}}$
$ = \dfrac{{67}}{{12}}$
So the required value of the given trigonometric equation (ii) is $\dfrac{{67}}{{12}}$ .
Therefore the option (C) is correct.
So this is the required answer.

Note: In the second method we can directly use the basic values of trigonometric ratios like the value of $\sin {60^ \circ } = \cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$ and $\sin {30^ \circ } = \cos {60^ \circ } = \dfrac{1}{2}$ and $\tan {45^ \circ } = 1$ . Substitution of these values directly into the given equation will give the answer. It is always advised to remember the trigonometric ratio values at some frequently used angles like ${30^ \circ },{45^ \circ },{60^ \circ },{90^ \circ }$ and ${120^ \circ }$ .