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

Evaluate arc⁡(tan(−3))\operatorname{arc} \left( {tan\left( { - \sqrt 3 } \right)} \right).

Explanation

Solution

Arc of a trigonometric function is the inverse of the trigonometric function. So first assume the given expression as any angle such as θ\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 arc⁡(tan(−3))\operatorname{arc} \left( {tan\left( { - \sqrt 3 } \right)} \right), that is the inverse of tan⁡(−3)\tan \left( { - \sqrt 3 } \right).
To do so first assume the given expression as an angle,
Let, tan⁡−1(−3)=θ{\tan ^{ - 1}}\left( { - \sqrt 3 } \right) = \theta
Taking tan⁡\tan on both sides, we get,
tan⁡(tan⁡−1(−3))=tan⁡θ\tan \left( {{{\tan }^{ - 1}}\left( { - \sqrt 3 } \right)} \right) = \tan \theta
⇒tan⁡θ=−3\Rightarrow \tan \theta = - \sqrt 3………(1)
Now
tan⁡π3=3\tan \dfrac{\pi }{3} = \sqrt 3
⇒−tan⁡π3=−3\Rightarrow - \tan \dfrac{\pi }{3} = - \sqrt 3
Using tan⁡(−θ)=−tan⁡θ\tan \left( { - \theta } \right) = - \tan \theta , we get
⇒tan⁡(−π3)=−3\Rightarrow \tan \left( { - \dfrac{\pi }{3}} \right) = - \sqrt 3 …………(2)
Comparing (1) and (2), we get,
tan⁡θ=tan⁡(−π3)\tan \theta = \tan \left( { - \dfrac{\pi }{3}} \right)
⇒θ=(−π3)\Rightarrow \theta = \left( { - \dfrac{\pi }{3}} \right)
Now −π2⩽θ⩽π2 - \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(−3){\tan ^{ - 1}}\left( { - \sqrt 3 } \right) is −π3 - \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{\tan ^{ - 1}}. To solve these problems memorise the values of sin, cos, tan, cosec, sec, cot for the angles 0,π6,π4,π3,π20,\dfrac{\pi }{6},\dfrac{\pi }{4},\dfrac{\pi }{3},\dfrac{\pi }{2}, for easy substitution of values.