Question

The ratio of \[\tan {60^0}\] and \[\cot {60^0}\] is

Answer 1

- Hint: The ratio of \[x\] and \[y\] is given by \[\dfrac{x}{y}\]. Then convert the terms of cot into tan by using simple trigonometric ratios. So, use this concept to reach the solution of the given problem.


Complete step-by-step solution -

The ratio of \[\tan {60^0}\] and \[\cot {60^0}\] is given by \[\dfrac{{\tan {{60}^0}}}{{\cot {{60}^0}}}\].
By using the formula \[\cot x = \dfrac{1}{{\tan x}}\], we have
\[
  \dfrac{{\tan {{60}^0}}}{{\cot {{60}^0}}} = \dfrac{{\tan {{60}^0}}}{{\dfrac{1}{{\tan {{60}^0}}}}} \\
  \dfrac{{\tan {{60}^0}}}{{\cot {{60}^0}}} = \tan {60^0} \times \tan {60^0} \\
  \dfrac{{\tan {{60}^0}}}{{\cot {{60}^0}}} = {\left( {\tan {{60}^0}} \right)^2} \\
  \dfrac{{\tan {{60}^0}}}{{\cot {{60}^0}}} = {\tan ^2}{60^0} \\
\]
We know that \[\tan {60^0} = \sqrt 3 \]
\[\dfrac{{\tan {{60}^0}}}{{\cot {{60}^0}}} = {\left( {\sqrt 3 } \right)^2} = 3\]
Thus, the ratio of \[\tan {60^0}\] and \[\cot {60^0}\] is 3.

Note: Remember all that the conversions of trigonometric ratios are the angles of the first quadrant only. We can solve this problem by another method which is given by
\[
  \dfrac{{\tan {{60}^0}}}{{\cot {{60}^0}}} = \dfrac{{\dfrac{{\sin {{60}^0}}}{{\cos {{60}^0}}}}}{{\dfrac{{\cos {{60}^0}}}{{\sin {{60}^0}}}}} \\
  \dfrac{{\tan {{60}^0}}}{{\cot {{60}^0}}} = \dfrac{{\sin {{60}^0}}}{{\cos {{60}^0}}} \times \dfrac{{\sin {{60}^0}}}{{\cos {{60}^0}}} \\
  \dfrac{{\tan {{60}^0}}}{{\cot {{60}^0}}} = {\left( {\dfrac{{\sin {{60}^0}}}{{\cos {{60}^0}}}} \right)^2} \\
  \dfrac{{\tan {{60}^0}}}{{\cot {{60}^0}}} = {\left( {\tan {{60}^0}} \right)^2} \\
  \therefore \dfrac{{\tan {{60}^0}}}{{\cot {{60}^0}}} = {\left( {\sqrt 3 } \right)^2} = 3 \\
\]