
How do you solve $\tan (x) = \sqrt 3 $?
Answer
497.1k+ views
Hint:In order to determine the value of the above question, take inverse of tangent on both sides of the equation to pull out x from inside of the tangent. Now we simply have to find the value of inverse of tangent of $\sqrt 3 $.The tangent function is positive in both 1 st and 3 rd quadrant so the answer exists for both the quadrant .Period of tangent is $\pi $,so frame your answer according to it
Complete step by step solution:
We are given a trigonometric expression $\tan (x) = \sqrt 3 $
$\tan (x) = \sqrt 3 $
Taking the inverse of tangent on both sides of the equation to pull out variable from inside the tangent ,we get
$
\Rightarrow {\tan ^{ - 1}}\left( {\tan (x)} \right) = {\tan ^{ - 1}}\left( {\sqrt 3 } \right) \\
\Rightarrow x = {\tan ^{ - 1}}\left( {\sqrt 3 } \right) \\
$
The value of ${\tan ^{ - 1}}\left( {\sqrt 3 } \right) = \dfrac{\pi }{3}$
Since, we know that the tangent function is always positive in the both first and third quadrants.
In order to determine the second solution of which tangent is equal to $\sqrt 3 $, we will add the reference angle $\pi $
$x = \pi + \dfrac{\pi }{3}$
Now simplifying the above,
$
x = \dfrac{{3\pi + \pi }}{3} \\
x = \dfrac{{4\pi }}{3} \\
$
$\tan (\theta + \pi ) = \dfrac{{\sin (\theta + \pi )}}{{\cos (\theta + \pi )}} = \dfrac{{ - \sin (\theta
)}}{{ - \cos (\theta )}} = \dfrac{{\sin (\theta )}}{{\cos (\theta )}} = \tan \left( \theta \right)$
So the period of $\tan \theta $ function is $\pi $,
$\therefore x = \dfrac{\pi }{3} + n\pi ,\dfrac{{4\pi }}{3} + n\pi $for any integer n
Therefore, solution of $\tan (x) = \sqrt 3 $is equal to $x = \dfrac{\pi }{3} + n\pi ,\dfrac{{4\pi }}{3} + n\pi $for any integer n.
Additional information:
1. In Mathematics the inverse trigonometric functions (every so often additionally called anti- trigonometric functions or cyclomatic function) are the reverse elements of the mathematical functions In particular, they are the inverses of the sine, cosine, tangent, cotangent, secant, and cosecant functions, and are utilized to get a point from any of the point's mathematical proportions.
Reverse trigonometric functions are generally utilized in designing, route, material science, and calculation.
2. In inverse trigonometric function, the domain are the ranges of corresponding trigonometric functions and the range are the domain of the corresponding trigonometric function.
3. Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus.
4. Periodic Function= A function $f(x)$ is said to be a periodic function if there exists a real number T > 0 such that $f(x + T) = f(x)$ for all x.
Note: 1.One must be careful while taking values from the trigonometric table and
cross-check at least once to avoid any error in the answer.
Complete step by step solution:
We are given a trigonometric expression $\tan (x) = \sqrt 3 $
$\tan (x) = \sqrt 3 $
Taking the inverse of tangent on both sides of the equation to pull out variable from inside the tangent ,we get
$
\Rightarrow {\tan ^{ - 1}}\left( {\tan (x)} \right) = {\tan ^{ - 1}}\left( {\sqrt 3 } \right) \\
\Rightarrow x = {\tan ^{ - 1}}\left( {\sqrt 3 } \right) \\
$
The value of ${\tan ^{ - 1}}\left( {\sqrt 3 } \right) = \dfrac{\pi }{3}$
Since, we know that the tangent function is always positive in the both first and third quadrants.
In order to determine the second solution of which tangent is equal to $\sqrt 3 $, we will add the reference angle $\pi $
$x = \pi + \dfrac{\pi }{3}$
Now simplifying the above,
$
x = \dfrac{{3\pi + \pi }}{3} \\
x = \dfrac{{4\pi }}{3} \\
$
$\tan (\theta + \pi ) = \dfrac{{\sin (\theta + \pi )}}{{\cos (\theta + \pi )}} = \dfrac{{ - \sin (\theta
)}}{{ - \cos (\theta )}} = \dfrac{{\sin (\theta )}}{{\cos (\theta )}} = \tan \left( \theta \right)$
So the period of $\tan \theta $ function is $\pi $,
$\therefore x = \dfrac{\pi }{3} + n\pi ,\dfrac{{4\pi }}{3} + n\pi $for any integer n
Therefore, solution of $\tan (x) = \sqrt 3 $is equal to $x = \dfrac{\pi }{3} + n\pi ,\dfrac{{4\pi }}{3} + n\pi $for any integer n.
Additional information:
1. In Mathematics the inverse trigonometric functions (every so often additionally called anti- trigonometric functions or cyclomatic function) are the reverse elements of the mathematical functions In particular, they are the inverses of the sine, cosine, tangent, cotangent, secant, and cosecant functions, and are utilized to get a point from any of the point's mathematical proportions.
Reverse trigonometric functions are generally utilized in designing, route, material science, and calculation.
2. In inverse trigonometric function, the domain are the ranges of corresponding trigonometric functions and the range are the domain of the corresponding trigonometric function.
3. Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus.
4. Periodic Function= A function $f(x)$ is said to be a periodic function if there exists a real number T > 0 such that $f(x + T) = f(x)$ for all x.
Note: 1.One must be careful while taking values from the trigonometric table and
cross-check at least once to avoid any error in the answer.
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