Question

The value of $\cos ec{60^ \circ }\cot {30^ \circ }\tan {60^ \circ }$ is equal to:
a) $2\sec {45^\circ }\cos {30^\circ }$
b) $2{\sec ^2}{45^\circ }\cos {30^\circ }$
c) $3\sin {60^ \circ }\sec {45^\circ }$
d) $3\sec {45^\circ }\cos {30^\circ }$

Answer 1

In this type of questions, we can also try by simply putting the values of trigonometric functions for different values of angles. But here we try to simplify $\cos ec{60^ \circ }\cot {30^ \circ }\tan {60^ \circ }$ to reach one of the options. Here, try to convert $\cos ec{60^ \circ }\cot {30^ \circ }\tan {60^ \circ }$ into a single trigonometric function. This will help us in reaching our answer.
Formula used: We have used the following functions here,
To convert trigonometric function between sin and cos, we use this identity,
$\sin \theta = \cos \left( {{{90}^ \circ } - \theta } \right)$
To convert trigonometric function between tan and cot, we use this identity,
$\cot \theta = \tan \left( {{{90}^ \circ } - \theta } \right)$
cot can also be converted in sin and cos function as
$\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$
Relation between sin and cosec function is given as,
$\dfrac{1}{{\sin \theta }} = \cos ec\theta$

Complete step-by-step solution:
So, the trigonometric function given to us is $\cos ec{60^ \circ }\cot {30^ \circ }\tan {60^ \circ }$.
First we convert tan function into cot. We know that $\cot \theta = \tan \left( {{{90}^ \circ } - \theta } \right)$. So,

$$\cos ec{60^ \circ }\cot {30^ \circ }\tan {60^ \circ } = \cos ec{60^ \circ }\cot {30^ \circ }\tan \left( {{{90}^ \circ } - {{60}^ \circ }} \right) \\\ \Rightarrow \cos ec{60^ \circ }\cot {30^ \circ }\tan {60^ \circ } = \cos ec{60^ \circ }\cot {30^ \circ }\cot {30^ \circ } \\\ \Rightarrow \cos ec{60^ \circ }\cot {30^ \circ }\tan {60^ \circ } = \cos ec{60^ \circ }{\cot ^2}{30^ \circ } \\\$$

As, $\cot \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$, we use this formula in above step and move ahead as,
$\cos ec{60^ \circ }\cot {30^ \circ }\tan {60^ \circ } = \cos ec{60^ \circ }\dfrac{{{{\cos }^2}{{30}^ \circ }}}{{{{\sin }^2}{{30}^ \circ }}}$
As we know that $\cos ec{60^ \circ } = \dfrac{1}{{\sin {{60}^\circ }}}$, so

$$\cos ec{60^ \circ }\cot {30^ \circ }\tan {60^ \circ } = \dfrac{1}{{\sin {{60}^ \circ }}}\dfrac{{{{\cos }^2}{{30}^ \circ }}}{{{{\sin }^2}{{30}^ \circ }}} \\\ \Rightarrow \cos ec{60^ \circ }\cot {30^ \circ }\tan {60^ \circ } = \dfrac{1}{{\cos {{30}^ \circ }}}\dfrac{{{{\cos }^2}{{30}^ \circ }}}{{{{\sin }^2}{{30}^ \circ }}} \\\$$

We know that one $\cos {30^ \circ }$is divided by another. We also write square of ${\sin ^2}{30^ \circ }$ separately,
$\cos ec{60^ \circ }\cot {30^ \circ }\tan {60^ \circ } = \dfrac{1}{{\sin {{30}^ \circ }}}\dfrac{{\cos {{30}^ \circ }}}{{\sin {{30}^ \circ }}}$
Putting the value of $\dfrac{1}{{\sin {{30}^ \circ }}} = 2$ and $\dfrac{1}{{\sin {{30}^ \circ }}} = \cos ec{30^ \circ }$ simultaneously, we get

$$\cos ec{60^ \circ }\cot {30^ \circ }\tan {60^ \circ } = 2\cos ec{30^ \circ }\cos {30^ \circ } \\\ \Rightarrow \cos ec{60^ \circ }\cot {30^ \circ }\tan {60^ \circ } = 2\sec {60^ \circ }\cos {30^ \circ } \\\ \Rightarrow \cos ec{60^ \circ }\cot {30^ \circ }\tan {60^ \circ } = 2 \times 2 \times \cos {30^ \circ } \\\ \Rightarrow \cos ec{60^ \circ }\cot {30^ \circ }\tan {60^ \circ } = 2{\left( {\sqrt 2 } \right)^2}\cos {30^ \circ } \\\ \Rightarrow \cos ec{60^ \circ }\cot {30^ \circ }\tan {60^ \circ } = 2{\sec ^2}45^\circ \cos {30^ \circ } \\\$$

This is equal to option b). Hence we have simplified the question upto a point where we have reached one of the options .

Note: It is to note that result is not the most simplified form of the given question. We have just converted the question expression into our desired form. This shows that any trigonometric function can be written in the form of another trigonometric function depending on our choice and need.