Question

How do you evaluate $\cot \left( \dfrac{7\pi }{6} \right)$ ?

Answer 1

Hint: For answering this question we need to evaluate the value of the given expression $\cot \left( \dfrac{7\pi }{6} \right)$ . We can observe that the angle is greater than $\pi $ , it means it will lie in the $3^{rd}$ quadrant. In the third quadrant, cotangent is positive.

Complete step by step solution:
Now considering from the question we have been asked to evaluate the value of the given expression $\cot \left( \dfrac{7\pi }{6} \right)$ .
We can observe that the angle in the given expression is greater than $\pi $ , then it means it will lie in the ${{3}^{rd}}$ quadrant.
In the third quadrant, the cotangent is positive.
This expression can be simplified and further written as $ \cot \left( \dfrac{7\pi }{6} \right)=\cot \left( \pi +\dfrac{\pi }{6} \right)$ .
By further simplifying this and writing it as $\cot \left( \dfrac{\pi }{6} \right)$ .
From the basic concepts of trigonometry we know that the value of $\cot \left( \dfrac{\pi }{6} \right)$ is $\sqrt{3}$ .

Therefore we can conclude that the value of the given trigonometric expression $\cot \left( \dfrac{7\pi }{6} \right)$ is given as $\sqrt{3}$ .

Note: While answering this question we should be sure with our concept and observations. If we observe the given expression correctly and apply our basic concepts then we will have an accurate result in a short span of time. We have four quadrants and eight trigonometric ratios. In the first quadrant, all trigonometric functions are positive. In the second quadrant, sine and cosecant trigonometric functions are positive and the remaining all are negative. In the third quadrant, tangent and cotangent trigonometric functions are positive and the remaining all are negative. In the fourth quadrant, cosecant and secant trigonometric functions are positive and the remaining all are negative. Similarly we can derive the values of any trigonometric expressions for example if we consider the trigonometric expression $\sin \left( \dfrac{7\pi }{6} \right)$ we will have its value as $ -\sin \left( \dfrac{\pi }{6} \right)=\dfrac{-1}{2}$.