Question

The ratio of \[\tan {{60}^{\circ }}\] and \[\cot {{60}^{\circ }}\] is

Answer 1

Hint:
 We are given \[\tan {{60}^{\circ }}\] and \[\cot {{60}^{\circ }}\] and we are asked to find the ratio of these two trigonometric ratios. We will first learn how these two ratios are related to each other then we use \[\tan \theta =\dfrac{1}{\cot \theta }\] or \[\cot \theta =\dfrac{1}{\tan \theta }\] to find the value of the ratio \[\dfrac{\tan {{60}^{\circ }}}{\cot {{60}^{\circ }}},\] as \[\tan {{60}^{\circ }}\] is \[\sqrt{3}.\] So, we get to put this value and solve further.

Complete step by step answer:
We are given that we have \[\tan {{60}^{\circ }}\] and \[\cot {{60}^{\circ }}\] and we have to find or evaluate the ratio of \[\tan {{60}^{\circ }}\] and \[\cot {{60}^{\circ }}.\] To find the ratio we will have to learn about these ratios. So, we will first learn about these ratios. Tan is the trigonometric ratio which is given as \[\tan \theta =\dfrac{\text{perpendicular}}{\text{base}}.......\left( i \right)\]
The cot is another trigonometric ratio which is defined as the ratio of the base to the perpendicular, that is \[\cot \theta =\dfrac{\text{base}}{\text{perpendicular}}.......\left( ii \right)\] where \[\theta \] is the angle.
We can observe that \[\cot \theta \] and \[\tan \theta \] are interrelated to each other.
Using (i) and (ii), we can see that,
\[\cot \theta =\dfrac{\text{base}}{\text{perpendicular}}\]
\[\Rightarrow \cot \theta =\dfrac{1}{\dfrac{\text{perpendicular}}{\text{base}}}\]
So, we get,
\[\Rightarrow \cot \theta =\dfrac{1}{\tan \theta }\]
Now, we have to find the value of the ratio of \[\tan {{60}^{\circ }}\] and \[\cot {{60}^{\circ }}.\] So \[\dfrac{\tan {{60}^{\circ }}}{\cot {{60}^{\circ }}}.\]
So, using the above relation, we can get,
\[\Rightarrow \dfrac{\tan {{60}^{\circ }}}{\cot {{60}^{\circ }}}=\dfrac{\tan {{60}^{\circ }}}{\dfrac{1}{\tan {{60}^{\circ }}}}\]
Simplifying, we will get,
\[\Rightarrow \dfrac{\tan {{60}^{\circ }}}{\cot {{60}^{\circ }}}=\dfrac{\tan {{60}^{\circ }}\times \tan {{60}^{\circ }}}{1}\]
\[\Rightarrow \dfrac{\tan {{60}^{\circ }}}{\cot {{60}^{\circ }}}=\dfrac{{{\left( \tan {{60}^{\circ }} \right)}^{2}}}{1}\]
As the value of \[\tan {{60}^{\circ }}\] is \[\sqrt{3}\] so we get,
\[\Rightarrow \dfrac{\tan {{60}^{\circ }}}{\cot {{60}^{\circ }}}=\dfrac{{{\left( \sqrt{3} \right)}^{2}}}{1}\]
\[\Rightarrow \dfrac{\tan {{60}^{\circ }}}{\cot {{60}^{\circ }}}=\dfrac{3}{1}\]
Therefore we get the required ratio as \[\dfrac{\tan {{60}^{\circ }}}{\cot {{60}^{\circ }}}=\dfrac{3}{1}.\]
\[\Rightarrow \tan {{60}^{\circ }}:\cot {{60}^{\circ }}=3:1\]

Note:
Another way to find the ratio is to put the value of \[\tan {{60}^{\circ }}\] and \[\cot {{60}^{\circ }}\] and then simplify. As \[\tan {{60}^{\circ }}=\sqrt{3}\] and \[\cot {{60}^{\circ }}=\dfrac{1}{\sqrt{3}}\] we can write,
\[\Rightarrow \dfrac{\tan {{60}^{\circ }}}{\cot {{60}^{\circ }}}=\dfrac{\sqrt{3}}{\dfrac{1}{\sqrt{3}}}\]
On simplifying, we get,
\[\Rightarrow \dfrac{\tan {{60}^{\circ }}}{\cot {{60}^{\circ }}}=\dfrac{\sqrt{3}\times \sqrt{3}}{1}\]
Further simplifying, we get,
\[\Rightarrow \dfrac{\tan {{60}^{\circ }}}{\cot {{60}^{\circ }}}=\dfrac{3}{1}\]
Therefore, the required ratio is
\[\Rightarrow \tan {{60}^{\circ }}:\cot {{60}^{\circ }}=3:1\]