Question

Why is the value of \[\tan 0 = 0 \] ?

Answer 1

Hint: In this question, we need to explain why the value of \[\tan 0^{o}\] is \[0\] . We can find the value of \[\tan 0^{o}\] by using trigonometric identities and ratios. The tangent is nothing but a ratio of the opposite side of a right angle to the adjacent side of the right angle. The basic trigonometric functions are sine , cosine and tangent. The values of \[\sin 0^{o}\] and \[\cos 0^{o}\] are used to find the value of \[\tan 0^{o}\] . With the help of these trigonometric functions , we can explain why the value of \[\tan 0^{o}\] is \[0\] .

Formula used :
\[\tan \theta = \dfrac{\sin \theta\ }{\cos \theta\ }\]
Trigonometry table :

Angle	\[0^{o}\]	\[30^{o}\]	\[45^{o}\]	\[60^{o}\]	\[90^{o}\]
Sine	\[0\]	\[\dfrac{1}{2}\]	\[\dfrac{1}{\sqrt{2}}\]	\[\dfrac{\sqrt{3}}{2}\]	\[1\]
Cosine	\[1\]	\[\dfrac{\sqrt{3}}{2}\]	\[\dfrac{1}{\sqrt{2}}\]	\[\dfrac{1}{2}\]	\[0\]

Complete step-by-step solution:
First we can find the value of \[\tan 0^{o}\]
We can find the value of \[\tan 0^{o}\] by using sine and cosine functions.
We know that,
\[\tan \theta\ = \dfrac{\sin \theta\ }{\cos \theta\ }\]
Here \[\theta = 0^{o}\] ,
Thus we get,
\[\tan 0^{o} = \dfrac{\sin 0^{o}}{\cos 0^{o}}\]
We know that \[\sin 0^{o}\] is \[0\] and \[\cos 0^{o}\] is \[1\] ,
By substituting the known values,
We get,
\[\tan 0^{o} = \dfrac{0}{1}\]
By dividing,
We get,
\[\tan 0^{o}= 0\]
Thus we get that the value of \[\tan 0^{o}\] is equal to \[0\] .
Since the value of \[\sin 0^{o}\] is \[0\] and \[\cos 0^{o}\] is \[1\],on dividing both we get the value of \[\tan 0^{o}\] as \[0\] .

Note: The concept used in this problem is trigonometric identities and ratios. Trigonometric identities are nothing but they involve trigonometric functions including variables and constants. The common technique used in this problem is the use of trigonometric functions. We need to note that \[\dfrac{0}{1}\] is \[0\] and \[\dfrac{1}{0}\] is \[\infty\] . In trigonometry , the tangent function is used to find the slope of a line .