Question & Answer
QUESTION

Find the value of the following:
$\dfrac{{5{{\sin }^2}{{30}^ \circ } + {{\cos }^2}{{45}^ \circ } - 4{{\tan }^2}{{30}^ \circ }}}{{2{{\sin }^2}{{30}^ \circ } + \cos {{30}^ \circ } + \tan {{45}^ \circ }}}$.

ANSWER Verified Verified
Hint: Put the value of $\sin {30^ \circ },\cos {30^ \circ },\tan {30^ \circ },\cos {45^ \circ },\tan {45^ \circ }$in the given expression-$\dfrac{{5{{\sin }^2}{{30}^ \circ } + {{\cos }^2}{{45}^ \circ } - 4{{\tan }^2}{{30}^ \circ }}}{{2{{\sin }^2}{{30}^ \circ } + \cos {{30}^ \circ } + \tan {{45}^ \circ }}}$, and then solve it to find the answer.

Complete step-by-step answer:
Given expression-
$\dfrac{{5{{\sin }^2}{{30}^ \circ } + {{\cos }^2}{{45}^ \circ } - 4{{\tan }^2}{{30}^ \circ }}}{{2{{\sin }^2}{{30}^ \circ } + \cos {{30}^ \circ } + \tan {{45}^ \circ }}}$
Now, we know the values of - \[\sin {30^ \circ } = \dfrac{1}{2},\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2},\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }},\tan {45^ \circ } = 1,\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}\].
Put these values in the expression given, we get-
$\dfrac{{5{{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^2} - 4{{\left( {\dfrac{1}{{\sqrt 3 }}} \right)}^2}}}{{2{{\left( {\dfrac{1}{2}} \right)}^2} + \dfrac{{\sqrt 3 }}{2} + 1}}$.
Solving it further,
$
  \dfrac{{5{{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^2} - 4{{\left( {\dfrac{1}{{\sqrt 3 }}} \right)}^2}}}{{2{{\left( {\dfrac{1}{2}} \right)}^2} + \dfrac{{\sqrt 3 }}{2} + 1}} \\
   = \dfrac{{5\left( {\dfrac{1}{4}} \right) + \dfrac{1}{2} - 4\left( {\dfrac{1}{3}} \right)}}{{2\left( {\dfrac{1}{4}} \right) + \dfrac{{\sqrt 3 }}{2} + 1}} \\
   = \dfrac{{\dfrac{5}{4} + \dfrac{1}{2} - \dfrac{4}{3}}}{{\dfrac{1}{2} + \dfrac{{\sqrt 3 }}{2} + 1}} = \dfrac{{\dfrac{{15 + 6 - 16}}{{12}}}}{{\dfrac{{1 + \sqrt 3 + 2}}{2}}} = \dfrac{5}{{12}} \times \dfrac{2}{{3 + \sqrt 3 }} = \dfrac{5}{6} \times \dfrac{1}{{3 + \sqrt 3 }} \\
 $
Multiplying and dividing by $3 - \sqrt 3 $, we get-
$\dfrac{5}{6} \times \dfrac{1}{{3 + \sqrt 3 }} \times \dfrac{{3 - \sqrt 3 }}{{3 - \sqrt 3 }} = \dfrac{{5(3 - \sqrt 3 )}}{{6({3^2} - {{(\sqrt 3 )}^2})}} = \dfrac{{5(3 - \sqrt 3 )}}{{6(9 - 3)}} = \dfrac{{5(3 - \sqrt 3 )}}{{36}}$.
Hence, the expression value is equal to $\dfrac{{5(3 - \sqrt 3 )}}{{36}}$.

Note: Whenever such types of questions appear, then always write down the expression given in the question and then put the values of-\[\sin {30^ \circ } = \dfrac{1}{2},\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2},\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }},\tan {45^ \circ } = 1,\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}\] in the expression and then step by step solve the equation, at last multiply with the conjugate of $3 + \sqrt 3 $, i.e., $3 - \sqrt 3 $ in numerator and denominator to obtain the final answer.