Prove that \[{\tan ^2}{30^ \circ } + 2\sin {60^ \circ } + \tan {45^ \circ } - \tan {60^ \circ } + {\cos ^2}{30^ \circ } = \dfrac{{25}}{{12}}\].

Hint: In this type of question simply put the value of angles and calculate them to get the desired answer. The value of
$\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}$
$\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2}$
$\tan {60^ \circ } = \sqrt 3 $
$\tan {45^ \circ } = 1$
$\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$.

Complete step-by-step answer:
Considering the left hand side of the given equation we get:-
 \[LHS = {\tan ^2}{30^ \circ } + 2\sin {60^ \circ } + \tan {45^ \circ } - \tan {60^ \circ } + {\cos ^2}{30^ \circ }\]
Now we know that,
$\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}$
$\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2}$
$\tan {60^ \circ } = \sqrt 3 $
$\tan {45^ \circ } = 1$
$\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$
Hence putting the respective values of angles we get:-
\[LHS = {\left( {\dfrac{1}{{\sqrt 3 }}} \right)^2} + 2\left( {\dfrac{{\sqrt 3 }}{2}} \right) + 1 - \sqrt 3 + \dfrac{3}{4}\]
Solving it further we get:-
We know that
\[{\left( {\sqrt 3 } \right)^2} = 3\]
Hence putting this value we get:-
\[LHS = \dfrac{1}{3} + \sqrt 3 + 1 - \sqrt 3 + \dfrac{3}{4}\]
Here the important part is $ + \sqrt 3 - \sqrt 3 = 0$
\[LHS = \dfrac{1}{3} + \dfrac{3}{4} + 1\]
Take the L.C.M of 3 and 4
\[
  LHS = \dfrac{{4\left( 1 \right) + 3\left( 3 \right)}}{{12}} + 1 \\
  LHS = \dfrac{{4 + 9}}{{12}} + 1 \\
  LHS = \dfrac{{13}}{{12}} + 1 \\
 \]
Now take the L.C.M of 12and 1
\[
  LHS = \dfrac{{13 + 12}}{{12}} \\
  LHS = \dfrac{{25}}{{12}} \\
 \]
 Hence \[LHS = RHS\]
Hence proved.

Note: Students may start solving these types of questions using identities however this approach is not correct instead they can solve the questions just by putting the values of angles and simply calculate and get answers.