Question

Evaluate \[\operatorname{arc} \left( {tan\left( { - \sqrt 3 } \right)} \right)\].

Answer 1

Hint: Arc of a trigonometric function is the inverse of the trigonometric function. So first assume the given expression as any angle such as \[\theta \], and then using trigonometric identities and transformations the required value can be found.

Complete step by step solution:
We have to find the value of \[\operatorname{arc} \left( {tan\left( { - \sqrt 3 } \right)} \right)\], that is the inverse of \[\tan \left( { - \sqrt 3 } \right)\].
To do so first assume the given expression as an angle,
Let, \[{\tan ^{ - 1}}\left( { - \sqrt 3 } \right) = \theta \]
Taking \[\tan \] on both sides, we get,
\[\tan \left( {{{\tan }^{ - 1}}\left( { - \sqrt 3 } \right)} \right) = \tan \theta \]
\[ \Rightarrow \tan \theta = - \sqrt 3 \]………(1)
Now
\[\tan \dfrac{\pi }{3} = \sqrt 3 \]
\[ \Rightarrow - \tan \dfrac{\pi }{3} = - \sqrt 3 \]
Using \[\tan \left( { - \theta } \right) = - \tan \theta \], we get
\[ \Rightarrow \tan \left( { - \dfrac{\pi }{3}} \right) = - \sqrt 3 \] …………(2)
Comparing (1) and (2), we get,
\[\tan \theta = \tan \left( { - \dfrac{\pi }{3}} \right)\]
\[ \Rightarrow \theta = \left( { - \dfrac{\pi }{3}} \right)\]
Now \[ - \dfrac{\pi }{2} \leqslant \theta \leqslant \dfrac{\pi }{2}\], that is lies within the range of the principle values of arctan.
Hence, the principal value of \[{\tan ^{ - 1}}\left( { - \sqrt 3 } \right)\] is \[ - \dfrac{\pi }{3}\].

Additional Information: Remember the STC or ASTC, that is the sin-tan-cos, or the arc sin-tan-cos rule for these problems. According to them, all trigonometric ratios are positive in first quadrant, only sin and cosec are positive in second quadrant, only tan and cot are positive in third quadrant and only cos and sec are positive in the fourth quadrant.

Note: Remember that arc tan means tan inverse and can be denoted as \[{\tan ^{ - 1}}\]. To solve these problems memorise the values of sin, cos, tan, cosec, sec, cot for the angles \[0,\dfrac{\pi }{6},\dfrac{\pi }{4},\dfrac{\pi }{3},\dfrac{\pi }{2}\], for easy substitution of values.