Answer

Verified

408.6k+ views

**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.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE

Which are the Top 10 Largest Countries of the World?

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

How do you graph the function fx 4x class 9 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths