Question

Find the value of $\dfrac{{\cos {{30}^o}\tan {{60}^o} + \sin {{60}^o}\cos {{60}^o}}}{{\cos ec{{30}^o}\sec {{60}^o} - \cos ec{{60}^o}\cot {{30}^o}}}$

Answer 1

In this particular question use the direct values of standard angles which is given as $\cos {30^o} = \dfrac{{\sqrt 3 }}{2}$,$\cos {60^o} = \dfrac{1}{2}$, $\tan {60^o} = \sqrt 3$, $\tan {30^o} = \dfrac{1}{{\sqrt 3 }}$, $\sin {60^o} = \dfrac{{\sqrt 3 }}{2}$, and $\sin {30^o} = \dfrac{1}{2}$, and also use the concept that cosec x = (1/sin x), sec x = (1/cos x), cot x = (1/cot x) so use these properties to reach the solution of the question.

Complete step-by-step answer:
Given trigonometric equation
$\dfrac{{\cos {{30}^o}\tan {{60}^o} + \sin {{60}^o}\cos {{60}^o}}}{{\cos ec{{30}^o}\sec {{60}^o} - \cos ec{{60}^o}\cot {{30}^o}}}$
So we have to find out the value of the above equation.
As we all know that cosec x = (1/sin x), sec x = (1/cos x), cot x = (1/cot x) so use these properties in the above equation we have,
$\Rightarrow \dfrac{{\cos {{30}^o}\tan {{60}^o} + \sin {{60}^o}\cos {{60}^o}}}{{\dfrac{1}{{\sin {{30}^o}}}\dfrac{1}{{\cos {{60}^o}}} - \dfrac{1}{{\sin {{60}^o}}}\dfrac{1}{{\tan {{30}^o}}}}}$
Now use the standard values of standard angles which is given as,
$\cos {30^o} = \dfrac{{\sqrt 3 }}{2}$,$\cos {60^o} = \dfrac{1}{2}$, $\tan {60^o} = \sqrt 3$, $\tan {30^o} = \dfrac{1}{{\sqrt 3 }}$, $\sin {60^o} = \dfrac{{\sqrt 3 }}{2}$, and $\sin {30^o} = \dfrac{1}{2}$ so substitute these values in the above equation we have,
$\Rightarrow \dfrac{{\dfrac{{\sqrt 3 }}{2}\left( {\sqrt 3 } \right) + \dfrac{{\sqrt 3 }}{2}\left( {\dfrac{1}{2}} \right)}}{{\left( {\dfrac{1}{{\dfrac{1}{2}}}} \right)\left( {\dfrac{1}{{\dfrac{1}{2}}}} \right) - \left( {\dfrac{1}{{\dfrac{{\sqrt 3 }}{2}}}} \right)\left( {\dfrac{1}{{\dfrac{1}{{\sqrt 3 }}}}} \right)}}$
Now simplify the above equation we have,
$\Rightarrow \dfrac{{\dfrac{{\sqrt 3 }}{2}\left( {\sqrt 3 } \right) + \dfrac{{\sqrt 3 }}{2}\left( {\dfrac{1}{2}} \right)}}{{\left( 2 \right)\left( 2 \right) - \left( {\dfrac{2}{{\sqrt 3 }}} \right)\left( {\sqrt 3 } \right)}}$
$\Rightarrow \dfrac{{\dfrac{3}{2} + \dfrac{{\sqrt 3 }}{4}}}{{4 - 2}} = \dfrac{{\dfrac{{6 + \sqrt 3 }}{4}}}{2} = \dfrac{{6 + \sqrt 3 }}{8}$
So this is the required answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the all the standard values of standard angles which are all stated above, these values is the key to the solution so directly substitute these values in the given equation after converting the given equation in terms of sine, cosine and tan as above we will get the required answer.