How do you solve \[\tan \theta = \sqrt 3 \] ?

Hint: The question is related to the inverse trigonometry topic. Here in this question to find the value of \[\tan \theta = \sqrt 3 \] , here we use inverse to find the value of \[\theta \] . To find the exact value we use the table of trigonometry ratios for standard angles and hence find the solution for the question.

Complete step-by-step answer:
The sine, cosine, tangent, cosecant, secant and cotangent are the trigonometry ratios of trigonometry. It is abbreviated as sin, cos, tan, cosec, sec and cot. Here in this question, we have \[\tan \theta = \sqrt 3 \] . By applying the inverse to the equation we solve the equation.
To find the value we use the table of trigonometry ratios for standard angles.
The table of tangent function for standard angles is given as


Now consider the given function
 \[\tan \theta = \sqrt 3 \]
So taking the inverse function we have
 \[ \Rightarrow \theta = {\tan ^{ - 1}}\left( {\sqrt 3 } \right)\]
From the table of tangent function for standard angles
 \[ \Rightarrow \theta = {60^ \circ }\]
This is in the form of degrees; let us convert into radians.
To convert the degree into radian we multiply the degree by \[\dfrac{\pi }{{180}}\]
Therefore, we have \[\theta = 60 \times \dfrac{\pi }{{180}}\]
On simplification we have
 \[ \Rightarrow \theta = \dfrac{\pi }{3}\]
Hence, we have solved the given trigonometric function.
Therefore, the value of \[\theta \] is \[\dfrac{\pi }{3}\] in radians and the value of \[\theta \] is \[{60^ \circ }\] in degree.

So, the correct answer is “The value of \[\theta \] is \[\dfrac{\pi }{3}\] in radians and the value of \[\theta \] is \[{60^ \circ }\] in degree”.

Note: The trigonometry and inverse trigonometry are inverse for each other. The inverse of a function is represented as the arc of the function or the function is raised by the power -1. For the trigonometry and the inverse trigonometry we need to know about the table of trigonometry ratios for the standard angles. In inverse trigonometry we have to take care of the domain and range of a function as well.