Question

How do I find the value of \[\cot \left( {\dfrac{\pi }{{12}}} \right)\]?

Answer 1

Hint: Trigonometric functions are those functions that tell us the relation between the three sides of a right-angled triangle. Sine, cosine, tangent, cosecant, secant and cotangent are the six types of trigonometric functions; sine, cosine and tangent are the main functions while cosecant, secant and cotangent are the reciprocal of sine, cosine and tangent respectively. Thus the given function can be converted in the form of tangent easily. First we find the value of tangent function then taking the reciprocal of tangent we get the cotangent value. Also we need to know the supplementary angle of sine.

Complete step-by-step solution:
Given, \[\cot \left( {\dfrac{\pi }{{12}}} \right)\].
We know that the
\[\cot \left( {\dfrac{\pi }{{12}}} \right) = \dfrac{1}{{\tan \left( {\dfrac{\pi }{{12}}} \right)}}\].
Now we find the value of \[\tan \left( {\dfrac{\pi }{{12}}} \right)\].
We can express \[\dfrac{\pi }{{12}} = \dfrac{\pi }{3} - \dfrac{\pi }{4}\]
Then we have
\[
 \Rightarrow \tan \left( {\dfrac{\pi }{{12}}} \right) = \tan \left( {\dfrac{\pi }{3} - \dfrac{\pi }{4}} \right) \\
    \Rightarrow \tan \left( {\dfrac{\pi }{3} - \dfrac{\pi }{4}} \right) \\
 \]
We know the difference formula for tangent that is \[\tan (A - B) = \dfrac{{\tan A - \tan B}}{{1 + \tan A.\tan B}}\]. Here \[A = \dfrac{\pi }{3}\]and\[B = \dfrac{\pi }{4}\].
\[ \Rightarrow \tan \left( {\dfrac{\pi }{3} - \dfrac{\pi }{4}} \right) = \dfrac{{\tan \dfrac{\pi }{3} - \tan \dfrac{\pi }{4}}}{{1 + \tan \dfrac{\pi }{3}.\tan \dfrac{\pi }{4}}}\]
\[ \Rightarrow \tan \left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{\tan \dfrac{\pi }{3} - \tan \dfrac{\pi }{4}}}{{1 + \tan \dfrac{\pi }{3}.\tan \dfrac{\pi }{4}}}\]
We know \[\tan \dfrac{\pi }{3} = \sqrt 3 \] and \[\tan \dfrac{\pi }{4} = 1\]. Substituting we have,
\[ \Rightarrow \dfrac{{\sqrt 3 - 1}}{{1 + \sqrt 3 .1}}\]
\[ \Rightarrow \dfrac{{\sqrt 3 - 1}}{{1 + \sqrt 3 }}\]
To simplify further we rationalize this
\[ \Rightarrow \dfrac{{\sqrt 3 - 1}}{{1 + \sqrt 3 }} \times \dfrac{{1 - \sqrt 3 }}{{1 - \sqrt 3 }}\]
\[ \Rightarrow \dfrac{{\left( {\sqrt 3 - 1} \right)\left( {1 - \sqrt 3 } \right)}}{{\left( {1 + \sqrt 3 } \right)\left( {1 - \sqrt 3 } \right)}}\]
Denominator is of the form \[{a^2} - {b^2} = (a + b)(a - b)\],
\[ \Rightarrow \dfrac{{\sqrt 3 \left( {1 - \sqrt 3 } \right) - 1\left( {1 - \sqrt 3 } \right)}}{{\left( {{1^2} - {{\left( {\sqrt 3 } \right)}^2}} \right)}}\]
\[ \Rightarrow \dfrac{{\sqrt 3 - {{\left( {\sqrt 3 } \right)}^2} - 1 + \sqrt 3 }}{{\left( {{1^2} - {{\left( {\sqrt 3 } \right)}^2}} \right)}}\]
Square and square root will cancel out,
\[ \Rightarrow \dfrac{{\sqrt 3 - 3 - 1 + \sqrt 3 }}{{\left( {1 - 3} \right)}}\]
\[ \Rightarrow \dfrac{{2\sqrt 3 - 4}}{{ - 2}}\]
Taking 2 common we have,
\[ \Rightarrow \dfrac{{2\left( {\sqrt 3 - 2} \right)}}{{ - 2}}\]
\[ \Rightarrow - \left( {\sqrt 3 - 2} \right)\]
\[ \Rightarrow 2 - \sqrt 3 \]
Thus we have \[\tan \left( {\dfrac{\pi }{{12}}} \right) = 2 - \sqrt 3 \].
Now we have,
\[\cot \left( {\dfrac{\pi }{{12}}} \right) = \dfrac{1}{{\tan \left( {\dfrac{\pi }{{12}}} \right)}}\]
\[ \Rightarrow \cot \left( {\dfrac{\pi }{{12}}} \right) = \dfrac{1}{{2 - \sqrt 3 }}\].

Note: Remember A graph is divided into four quadrants, all the trigonometric functions are positive in the first quadrant, all the trigonometric functions are negative in the second quadrant except sine and cosine functions, tangent and cotangent are positive in the third quadrant while all others are negative and similarly all the trigonometric functions are negative in the fourth quadrant except cosine and secant.