Question

How do you find the exact value of tan 300?

Answer 1

Hint: The given question is the trigonometric expression and in order to solve this solve we have to use the properties of trigonometric functions. First we need to remove the full rotation of \[2\pi \] until the angle is between 0 to \[2\pi \]. Then using the trigonometric ratios table, we will find the exact value of the given expression.

Formula used:
\[\pi \] radian = \[{{180}^{0}}\]
To find the tangent of any angle we need to just divide the sine and cosine of the same angle:\[\tan \theta =\dfrac{\sin \theta }{\cos \theta }=\dfrac{perpendicular}{base}\]

Trigonometric ratio table used to find the sine and cosine of the angle:

Angles(in degrees)	\[\sin \theta \]	\[\cos \theta \]
\[{{0}^{0}}\]	0	1
\[{{30}^{0}}\]	\[\dfrac{1}{2}\]	\[\dfrac{\sqrt{3}}{2}\]
\[{{45}^{0}}\]	\[\dfrac{1}{\sqrt{2}}\]	\[\dfrac{1}{\sqrt{2}}\]
\[{{60}^{0}}\]	\[\dfrac{\sqrt{3}}{2}\]	\[\dfrac{1}{2}\]
\[{{90}^{0}}\]	1	0

\[\]

Complete step by step answer:
We have given that,
\[\Rightarrow \tan \left( {{300}^{0}} \right)\]
As we know that;
\[\Rightarrow {{300}^{0}}={{360}^{0}}-{{60}^{0}}\]
And,
\[\Rightarrow \pi ={{180}^{0}}\]Then \[2\pi ={{360}^{0}}\]
\[\Rightarrow {{60}^{0}}=\dfrac{60}{180}\times \pi =\dfrac{\pi }{3}\]
Therefore,
\[\Rightarrow {{300}^{0}}={{360}^{0}}-{{60}^{0}}\]
Equivalently in radian, we have
\[\Rightarrow {{300}^{0}}=2\pi -\dfrac{\pi }{3}\]
Thus,
\[\Rightarrow \tan \left( {{300}^{0}} \right)=\tan \left( 2\pi -\dfrac{\pi }{3} \right)\]
On trigonometric unit circle;
\[\Rightarrow \tan \left( 2\pi -\dfrac{\pi }{3} \right)=\tan \left( -\dfrac{\pi }{3} \right)\]
Tangent function is an odd function,
Thus,
\[\Rightarrow \tan \left( -\dfrac{\pi }{3} \right)=-\tan \dfrac{\pi }{3}\]
Therefore,
We have
\[\Rightarrow -\tan \dfrac{\pi }{3}\]
Now the next step is to find the value of \[\dfrac{\pi }{3}\],,
As we know that,
\[\pi \] radian = \[{{180}^{0}}\]
Therefore, \[\dfrac{\pi }{3}=\dfrac{{{180}^{0}}}{3}={{60}^{0}}\]
Now putting \[\dfrac{\pi }{3}={{60}^{0}}\]
\[\tan \theta =\dfrac{\sin \theta }{\cos \theta }=\dfrac{perpendicular}{base}\]
\[-\tan {{60}^{0}}=-\left( \dfrac{\sin {{60}^{0}}}{\cos {{60}^{0}}} \right)\]
From the trigonometric table we know the value of
\[\sin {{60}^{0}}=\dfrac{\sqrt{3}}{2}\] and \[\cos {{60}^{0}}=\dfrac{1}{2}\]
\[-\tan {{60}^{0}}=-\left( \dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}} \right)=-\dfrac{\sqrt{3}}{1}=-\sqrt{3}\]
Therefore, the value of \[-\tan \dfrac{\pi }{3}=-\sqrt{3}\]
Thus,
. \[\Rightarrow \tan \left( {{300}^{0}} \right)=-\sqrt{3}\]
Hence, it is the required answer.

Note: One must be careful while noted down the values from the trigonometric table to avoid any error in the answer. We must know the basic value of sine and cosine of the angles like \[{{0}^{0}}\], \[{{30}^{0}}\], \[{{60}^{0}}\], \[{{90}^{0}}\] etc. Whenever we get this type of problem, first convert the radians to degrees to make the process of solving the question easier. The sine, cosine and the tangent are the three basic functions in introduction to trigonometry which shows the relation between all the sides of the triangles.