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Question: How do you find the exact value of \({\tan ^{ - 1}}\left( {\dfrac{{ - \sqrt 3 }}{3}} \right)\)?...

How do you find the exact value of tan1(33){\tan ^{ - 1}}\left( {\dfrac{{ - \sqrt 3 }}{3}} \right)?

Explanation

Solution

First see whether the value of θ\theta is known or not if not, then try to solve, so that we get the known value of θ\theta . And then check whether we have a proper sign in the number or not, if not then try to convert it into another angle so that the value matched with the θ\theta value.

Complete step by step answer:
To solve this equation, first check whether we know the value of θ\theta or not. In this problem, it seems we have to simplify more to get the solution. So, first let us consider the bracketed term,
(33)\left( {\dfrac{{ - \sqrt 3 }}{3}} \right) == (33×3)=(13)\left( {\dfrac{{ - \sqrt 3 }}{{\sqrt 3 \times \sqrt 3 }}} \right) = \left( {\dfrac{{ - 1}}{{\sqrt 3 }}} \right)
Now find the value of θ\theta , for which the value is (13)\left( {\dfrac{{ - 1}}{{\sqrt 3 }}} \right) and the value of θ\theta is in negative value. Andtanθ\tan \theta is negative in the first, second and fourth quadrant. We know the value for tan(π6)\tan \left( {\dfrac{\pi }{6}} \right) is (13)\left( {\dfrac{1}{{\sqrt 3 }}} \right). And the solution will be θ=π6\theta = \dfrac{\pi }{6}, but the value of tanπ6\tan \dfrac{\pi }{6} is equal to (13)\left( {\dfrac{1}{{\sqrt 3 }}} \right). But we have (13)\left( {\dfrac{{ - 1}}{{\sqrt 3 }}} \right), the angle cannot be negative. So, we are considering tan(πθ)=tanθ\tan \left( {\pi - \theta } \right) = - \tan \theta , substituting the value of θ\theta and we get,
tan(ππ6)=tan(5π6)\tan \left( {\pi - \dfrac{\pi }{6}} \right) = \tan \left( {\dfrac{{5\pi }}{6}} \right)
And the value of θ\theta will be equal to 5π6\dfrac{{5\pi }}{6}.
Additional information: Here, 180{180^ \circ } is expressed as π\pi because the circumference of a triangle is 2πr2\pi r.

Note: To solve this problem we must know the trigonometric ratios for related triangles. Because this is where many students will struggle. And we must know the value of sinθ,cosθ,tanθ,cosecθ,secθ,\sin \theta ,\cos \theta ,\tan \theta ,\cos ec\theta ,\sec \theta , and cotθ\cot \theta for the values 0,30,45,60,{0^ \circ },{30^ \circ },{45^ \circ },{60^ \circ }, and 90{90^ \circ }. Knowing these values, we can solve any problem by converting them into known values.