Question

Find the value of the trigonometric function: $\tan \left( \dfrac{19\pi }{3} \right)$.

Answer 1

Hint: In this question, we can use the concept that tangent and cotangent ratios of an angle θ and of the sum of a multiple of π and θ \[\left( n\times \pi +\theta \right)\] are equal. So, here we may convert $\tan \left( \dfrac{19\pi }{3} \right)$ into the form of $\tan \left( n\times \pi +\theta \right)$ where n is any integer, and then equalize it to $\tan \theta $ and get our required answer.

Complete step-by-step answer:
In this given question, we are asked to find the value of the trigonometric function: $\tan

\left( \dfrac{19\pi }{3} \right)$.

As we know, all trigonometric ratios of an angle θ and of the sum of multiples of 2π and θ

$\left( n\times 2\pi +\theta \right)$ are equal. Statement……. (1.1)

So, here we can convert the given trigonometric ratio of $\tan \left( \dfrac{19\pi }{3} \right)$

into the form of $\tan \left( n\times 2\pi +\theta \right)$ where n is any integer, that is $\tan

\left( 3\times 2\pi +\dfrac{\pi }{3} \right)$.

It gives us $\tan \left( \dfrac{19\pi }{3} \right)=\tan \left( 3\times 2\pi +\dfrac{\pi }{3}

\right)..........(1.1)$

Now, as per statement 1.1 we can write equation 1.1 as:

$\tan \left( \dfrac{19\pi }{3} \right)=\tan \left( 3\times 2\pi +\dfrac{\pi }{3} \right)=\tan \left(

\dfrac{\pi }{3} \right)..........(1.2)$

Now, we know that the value of $\left( \dfrac{\pi }{3} \right)$ corresponds to ${{60}^{\circ

}}$.

So, $\tan \left( \dfrac{\pi }{3} \right)=\tan {{60}^{\circ }}=\sqrt{3}..........(1.3)$

Hence, from equation 1.3, we get the value of $\tan \left( \dfrac{\pi }{3} \right)=\sqrt{3}$.

So, from equation 1.1 and 1.2,we get $\tan \left( \dfrac{19\pi }{3} \right)=\tan \left(

\dfrac{\pi }{3} \right)=\sqrt{3}..........(1.4)$.

Therefore, from equation 1.4 we have got our answer to the question as the value of $\tan

\left( \dfrac{19\pi }{3} \right)$as $\sqrt{3}$.


Note: In this question, we could also have used the concept that tangent and cotangent ratios of an angle θ and of the sum of a multiple of π and θ \[\left( n\times \pi +\theta \right)\] are equal, in order to arrive at the same answer we have got here.