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# The ratio of $\tan {{60}^{\circ }}$ and $\cot {{60}^{\circ }}$ is

Last updated date: 13th Sep 2024
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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
$\Rightarrow \tan {{60}^{\circ }}:\cot {{60}^{\circ }}=3:1$