Question

The value of $2{\cot ^2}\left( {\dfrac{\pi }{6}} \right) + 4{\tan ^2}\left( {\dfrac{\pi }{6}} \right) - 3cosec\dfrac{\pi }{6}$ is
$1){\text{ 2}}$
$2){\text{ }}4$
$3)\;\dfrac{4}{3}$
$4){\text{ }}\dfrac{3}{4}$

Answer 1

Hint : We have to find the value of the given trigonometric expression $2{\cot ^2}\left( {\dfrac{\pi }{6}} \right) + 4{\tan ^2}\left( {\dfrac{\pi }{6}} \right) - 3cosec\dfrac{\pi }{6}$ . We solve this question by substituting the respective values of the trigonometric functions in their respective expression . We already know the values of trigonometric functions for different values or different angles . So by substituting the values of trigonometric functions for the given value of angle we can evaluate the value of the given trigonometric expression $2{\cot ^2}\left( {\dfrac{\pi }{6}} \right) + 4{\tan ^2}\left( {\dfrac{\pi }{6}} \right) - 3cosec\dfrac{\pi }{6}$.

Complete step-by-step answer :
All the trigonometric functions are classified into two categories or types as either sine function or cosine function . All the functions which lie in the category of sine functions are sin , cosec and tan functions on the other hand the functions which lie in the category of cosine functions are cos , sec and cot functions . The trigonometric functions are classified into these two categories on the basis of their property which is stated as : when the value of angle is substituted by the negative value of the angle then we get the negative value for the functions in the sine family and a positive value for the functions in the cosine family .
Given :
We have to find the value of $\cot \left( {\dfrac{\pi }{6}} \right),cosec\left( {\dfrac{\pi }{6}} \right)$ and $\tan \left( {\dfrac{\pi }{6}} \right)$to solve the question .
From the trigonometric values of common angles, we know that $\dfrac{\pi }{6} = {30^ \circ }$.
Also we know that $cot{\text{ }}x{\text{ }} = \dfrac{1}{{tan{\text{ }}x}}$
so , we get
$\cot \left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{{\tan \left( {\dfrac{\pi }{6}} \right)}} = \dfrac{1}{{\dfrac{{\sqrt 3 }}{3}}} = \dfrac{3}{{\sqrt 3 }}$
And,
$\tan \left( {\dfrac{\pi }{6}} \right) = \dfrac{{\sqrt 3 }}{3}\& cosec\left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{{\sin \left( {\dfrac{\pi }{6}} \right)}} = \dfrac{1}{{\dfrac{1}{2}}} = 2$
As , $cosec{\text{ }}x = \dfrac{1}{{sin{\text{ }}x}}$
So there by substituting the values in the given question,
$2{\cot ^2}\left( {\dfrac{\pi }{6}} \right) + 4{\tan ^2}\left( {\dfrac{\pi }{6}} \right) - 3cosec\dfrac{\pi }{6}$
Substituting the values from the trigonometric values,
$=2 \times {\left( {\dfrac{3}{{\sqrt 3 }}} \right)^2} + 4 \times {\left( {\dfrac{{\sqrt 3 }}{3}} \right)^2} - 3 \times 2$
$= 2 \times \dfrac{9}{3} + 4 \times \dfrac{3}{9} - 6 = 6 + \dfrac{4}{3} - 6 = \dfrac{4}{3}$
Thus , the value of $2{\cot ^2}\left( {\dfrac{\pi }{6}} \right) + 4{\tan ^2}\left( {\dfrac{\pi }{6}} \right) - 3cosec\dfrac{\pi }{6}$ is $\dfrac{4}{3}$
Hence , the correct option is$(3)$.
So, the correct answer is “Option 3”.

Note : Whenever we are solving the trigonometric functions, we can simplify the equations by converting all the functions to sine and cosine. The various trigonometry formulas for the conversion are given as :
$sin{\text{ }}x{\text{ }} = {\text{ }}cos{\text{ }}\left( {90^\circ {\text{ }}-{\text{ }}x} \right)$
$cos{\text{ }}x{\text{ }} = {\text{ }}sin{\text{ }}\left( {90^\circ {\text{ }}-{\text{ }}x} \right)$
$tan{\text{ }}x{\text{ }} = {\text{ }}cot{\text{ }}\left( {90^\circ {\text{ }}-{\text{ }}x} \right)$
$cot{\text{ }}x{\text{ }} = {\text{ }}tan{\text{ }}\left( {90^\circ {\text{ }}-{\text{ }}x} \right)$
$sec{\text{ }}x{\text{ }} = {\text{ }}cosec{\text{ }}\left( {90^\circ {\text{ }}-{\text{ }}x} \right)$
$cosec{\text{ }}x{\text{ }} = {\text{ }}sec{\text{ }}\left( {90^\circ {\text{ }}-{\text{ }}x} \right)$