Solveeit Logo

Question

Question: How do you evaluate \(\tan \dfrac{{3\pi }}{2}\)?...

How do you evaluate tan3π2\tan \dfrac{{3\pi }}{2}?

Explanation

Solution

Here we are asked to find the tangent of an angle. We have to remember that a tangent function is a periodic function with periodicity π\pi . Therefore we can write,
tan(π+x)=tanx\tan \left( {\pi + x} \right) = \tan x , where 0xπ20 \leqslant x \leqslant \dfrac{\pi }{2}

Formula used:
tan(π+x)=tanx\tan \left( {\pi + x} \right) = \tan x

Complete step-by-step answer:
We are asked to evaluate tan3π2\tan \dfrac{{3\pi }}{2}. The angle is 3π2radians\dfrac{{3\pi }}{2}\,radians or 270270^\circ .
Since we are making use of the periodicity of tangent function, let us split 3π2\dfrac{{3\pi }}{2} as follows:
3π2=π2+π2+π2\dfrac{{3\pi }}{2} = \dfrac{\pi }{2} + \dfrac{\pi }{2} + \dfrac{\pi }{2}
3π2=π+π2\Rightarrow \dfrac{{3\pi }}{2} = \pi + \dfrac{\pi }{2}
Therefore, we can write question as
tan3π2=tan(π+π2)\tan \dfrac{{3\pi }}{2} = \tan \left( {\pi + \dfrac{\pi }{2}} \right)
tan(π+π2)=tanπ2\tan \left( {\pi + \dfrac{\pi }{2}} \right) = \tan \dfrac{\pi }{2}
That is,
tan3π2=tanπ2\Rightarrow \tan \dfrac{{3\pi }}{2} = \tan \dfrac{\pi }{2}
Now, we know that, tanθ=sinθcosθ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} , θ\theta is any angle.
Therefore, we have to find sinπ2,cosπ2\sin \dfrac{\pi }{2},\cos \dfrac{\pi }{2}.
Consider a right angled triangle, ΔAOB\Delta AOB , right angled at O\angle O.

From definition of sine, we have
sin(AOB)=oppositesidehypotenuse=ABAB=1\sin \left( {\angle AOB} \right) = \dfrac{{opposite\,side}}{{hypotenuse}} = \dfrac{{AB}}{{AB}} = 1
Now, consider B\angle B . If we bring the vertex BB along the side OBOB so that BB coincides with OO , we then will have BC=0BC = 0 .
Now from definition of cosine, we have
cos(AOB)=adjacentsidehypotenuse=BCAB=01=0\cos \left( {\angle AOB} \right) = \dfrac{{adjacent\,side}}{{hypotenuse}} = \dfrac{{BC}}{{AB}} = \dfrac{0}{1} = 0
Therefore, we have
tanπ2=sinπ2cosπ2=sin(AOB)cos(AOB)=10\tan \dfrac{\pi }{2} = \dfrac{{\sin \dfrac{\pi }{2}}}{{\cos \dfrac{\pi }{2}}} = \dfrac{{\sin \left( {\angle AOB} \right)}}{{\cos \left( {\angle AOB} \right)}} = \dfrac{1}{0}
We know that 10\dfrac{1}{0} is undefined.
That is, tanπ2=undefined\tan \dfrac{\pi }{2} = undefined
Therefore, going back to our original problem, we have,
tan3π2=tanπ2=undefined\tan \dfrac{{3\pi }}{2} = \tan \dfrac{\pi }{2} = undefined
tan(3π2)=undefined\Rightarrow \tan \left( {\dfrac{{3\pi }}{2}} \right) = undefined.

Note: The same result can be obtained graphically by plotting the graph of tangent function.The trigonometric function takes in angles as their input and gives out the ratio of sides of a triangle. Thus, these provide a link between angles and sides.