Question

Evaluate the following-

$\dfrac{{5{{\sin }^2}{{30}^ \circ } + {{\cos }^2}{{45}^ \circ } - 4{{\tan }^2}{{30}^ \circ }}}{{2\sin {{30}^ \circ }\cos {{30}^ \circ } + \tan {{45}^ \circ }}}$

Answer 1

Use the values of given trigonometric identities which are given as-

$\sin {30^ \circ } = \dfrac{1}{2}$ ,$\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$ , $\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$,$\tan {45^ \circ } = 1$ and $\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}$ . Put them in the given function and solve the values to get the answer.

Complete step-by-step answer:

We have to evaluate the given function

$\Rightarrow \dfrac{{5{{\sin }^2}{{30}^ \circ } + {{\cos }^2}{{45}^ \circ } - 4{{\tan }^2}{{30}^ \circ }}}{{2\sin {{30}^ \circ }\cos {{30}^ \circ } + \tan {{45}^ \circ }}}$

We know the values of trigonometric identities are given as-

$\sin {30^ \circ } = \dfrac{1}{2}$ ,$\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$ , $\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$,$\tan {45^ \circ } = 1$ and $\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}$

On using the values of these identities in the formula we get,

$\Rightarrow \dfrac{{5{{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^2} - 4 \times {{\left( {\dfrac{1}{{\sqrt 3 }}} \right)}^2}}}{{2\left( {\dfrac{1}{2}} \right)\left( {\dfrac{{\sqrt 3 }}{2}} \right) + 1}}$

On opening the brackets we get,

$\Rightarrow \dfrac{{5 \times \dfrac{1}{4} + \dfrac{1}{2} - 4 \times \dfrac{1}{3}}}{{\dfrac{{2 \times \sqrt 3 }}{{2 \times 2}} + 1}}$

On multiplying the required values we get,

$\Rightarrow \dfrac{{\dfrac{5}{4} + \dfrac{1}{2} - \dfrac{4}{3}}}{{\dfrac{{\sqrt 3 }}{2} + 1}}$

On Taking LCM on numerator and denominator we get,

$\Rightarrow \dfrac{{\dfrac{{\left( {5 \times 3} \right) + 6 - \left( {4 \times 3} \right)}}{{12}}}}{{\dfrac{{\sqrt 3 + 2}}{2}}}$

On solving the numerator we get,

$\Rightarrow \dfrac{{\dfrac{{15 + 6 - 16}}{{12}}}}{{\dfrac{{\sqrt 3 + 2}}{2}}}$

On simplifying we get,

$\Rightarrow \dfrac{{\dfrac{5}{{12}}}}{{\dfrac{{\sqrt 3 + 2}}{2}}}$

Now we can write the given function as-

$\Rightarrow \dfrac{5}{{12}} \times \dfrac{2}{{\sqrt 3 + 2}}$

On cancelling the terms which can be cancelled we get,

$\Rightarrow \dfrac{5}{6} \times \dfrac{1}{{\sqrt 3 + 2}}$

Then we get,

Answer$\Rightarrow$ $\dfrac{5}{{6\left( {2 + \sqrt 3 } \right)}}$

Note: You can also solve it further by rationalizing but it is not required. We can rationalize it by multiplying $2 - \sqrt 3$ on the numerator and denominator.

$\Rightarrow \dfrac{{5\left( {2 - \sqrt 3 } \right)}}{{6\left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)}}$

On using the identity $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$ we get,

$\Rightarrow \dfrac{{5\left( {2 - \sqrt 3 } \right)}}{{6\left( {{2^2} - {{\left( {\sqrt 3 } \right)}^2}} \right)}}$

On solving we get,

$\Rightarrow \dfrac{{5\left( {2 - \sqrt 3 } \right)}}{{6\left( {4 - 3} \right)}}$

On simplifying we get,

$\Rightarrow \dfrac{{5\left( {2 - \sqrt 3 } \right)}}{6}$

This can also be the answer of the given question.