Question

How do I evaluate \[\tan \left( \dfrac{\pi }{3} \right)\] without using a calculator?

Answer 1

Hint: To solve this equation, we need to know the relationship between the \[\tan x,\sin x\And \cos x\], which states that, \[\tan x=\dfrac{\sin x}{\cos x}\]. We will use this relation to find the value of \[\tan \left( \dfrac{\pi }{3} \right)\]. Also, we should know the values of \[\sin \left( \dfrac{\pi }{3} \right)\And \cos \left( \dfrac{\pi }{3} \right)\].

Complete step by step answer:
We know that the trigonometric ratios \[\tan x,\sin x\And \cos x\] are related to each other in the following way, \[\tan x=\dfrac{\sin x}{\cos x}\]. As \[\dfrac{\pi }{3}\] is a special angel, we know the values of \[\sin \left( \dfrac{\pi }{3} \right)\And \cos \left( \dfrac{\pi }{3} \right)\]. The value of \[\sin \left( \dfrac{\pi }{3} \right)\] equals \[\dfrac{\sqrt{3}}{2}\], and the value of \[\cos \left( \dfrac{\pi }{3} \right)\] is \[\dfrac{1}{2}\].
Using the relationship between the ratios, and these values, we can find the value of \[\tan \left( \dfrac{\pi }{3} \right)\] as follows
\[\tan x=\dfrac{\sin x}{\cos x}\]
Substituting \[x=\dfrac{\pi }{3}\], we get
\[\Rightarrow \tan \left( \dfrac{\pi }{3} \right)=\dfrac{\sin \left( \dfrac{\pi }{3} \right)}{\cos \left( \dfrac{\pi }{3} \right)}\]
Substituting the values of \[\sin \left( \dfrac{\pi }{3} \right)\And \cos \left( \dfrac{\pi }{3} \right)\] in the above equation, we get
\[\Rightarrow \tan \left( \dfrac{\pi }{3} \right)=\dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}}\]
As the fraction in the numerator and the fraction in the denominator have the same denominator, we can cancel it out, by doing this we get
\[\Rightarrow \tan \left( \dfrac{\pi }{3} \right)=\sqrt{3}\]
Thus, we get the value of \[\tan \left( \dfrac{\pi }{3} \right)\].

Note:
We can also find the value of \[\tan \left( \dfrac{\pi }{3} \right)\], if we know the value of the tangent of its half, that is the value of \[\tan \left( \dfrac{\pi }{6} \right)\]. The value of \[\tan \left( \dfrac{\pi }{6} \right)\] is \[\dfrac{1}{\sqrt{3}}\].
We know the formula for \[\tan (2x)\] is \[\dfrac{2\tan x}{1-{{\tan }^{2}}x}\]. As \[\dfrac{\pi }{3}\] is twice of the \[\dfrac{\pi }{6}\], we can use this formula to calculate the value of \[\tan \left( \dfrac{\pi }{3} \right)\], using the value of \[\tan \left( \dfrac{\pi }{6} \right)\], as follows
\[\tan (2x)=\dfrac{2\tan x}{1-{{\tan }^{2}}x}\]
Substitute \[x=\dfrac{\pi }{6}\] in the above formula, we get
\[\begin{align}
  & \Rightarrow \tan \left( 2\left( \dfrac{\pi }{6} \right) \right)=\dfrac{2\tan \left( \dfrac{\pi }{6} \right)}{1-{{\tan }^{2}}\left( \dfrac{\pi }{6} \right)} \\
 & \Rightarrow \tan \left( \dfrac{\pi }{3} \right)=\dfrac{2\tan \left( \dfrac{\pi }{6} \right)}{1-{{\tan }^{2}}\left( \dfrac{\pi }{6} \right)} \\
\end{align}\]
Substituting the value of \[\tan \left( \dfrac{\pi }{6} \right)\] as \[\dfrac{1}{\sqrt{3}}\] in the above formula, we get
\[\Rightarrow \tan \left( \dfrac{\pi }{3} \right)=\dfrac{2\left( \dfrac{1}{\sqrt{3}} \right)}{1-{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}}\]
\[\Rightarrow \tan \left( \dfrac{\pi }{3} \right)=\dfrac{\dfrac{2}{\sqrt{3}}}{1-\dfrac{1}{3}}=\dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{2}{3}}\]
To make the denominator of the fraction in numerator and denominator same, we multiply and divide the fraction in the numerator by \[\sqrt{3}\], we get
\[\Rightarrow \tan \left( \dfrac{\pi }{3} \right)=\dfrac{\dfrac{2}{\sqrt{3}}\times \dfrac{\sqrt{3}}{\sqrt{3}}}{\dfrac{2}{3}}=\dfrac{\dfrac{2\sqrt{3}}{3}}{\dfrac{2}{3}}\]
Canceling the common factors from numerator and denominator, we get
\[\Rightarrow \tan \left( \dfrac{\pi }{3} \right)=\sqrt{3}\]