Question

Evaluate the following: ${\tan ^{ - 1}}\left( {\tan 4} \right)$

Answer 1

Hint : To solve this problem, we need to understand the concept of inverse of trigonometric functions. The inverse trigonometric functions are also called Arc functions. Inverse Trigonometric Functions are defined in a certain interval under constrained domain. We will use the interval of the tangent function.

Complete step-by-step answer :
We will first see the range of inverse tangents to solve this problem. For all real numbers, the range of inverse tangent is $- \dfrac{\pi }{2} \leqslant \theta \leqslant \dfrac{\pi }{2}$ which means that ${\tan ^{ - 1}}\left( {\tan \theta } \right) = \theta$ if $\theta \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$ .
Here, we are given that ${\tan ^{ - 1}}\left( {\tan 4} \right)$ .
But here, $\theta = 4$ , which does not belong to the range of inverse tangent which is $- \dfrac{\pi }{2} \leqslant \theta \leqslant \dfrac{\pi }{2}$ .
We also know that $\tan (\pi - \theta ) = - \tan \theta$ .
Therefore, we can say that $\tan (\theta - \pi ) = \tan \theta$
Using this principle, we can say that $\tan (4 - \pi ) = \tan 4$
So, we can rewrite the given term as ${\tan ^{ - 1}}\left( {\tan \left( {4 - \pi } \right)} \right)$
Now, $4 - \pi$ belongs to the range of inverse tangent $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$ .
Therefore, we can write that ${\tan ^{ - 1}}\left( {\tan \left( {4 - \pi } \right)} \right) = 4 - \pi$
But, we have seen that $\tan (4 - \pi ) = \tan 4$ which means that ${\tan ^{ - 1}}\left( {\tan 4} \right) = 4 - \pi$
Thus, by evaluating ${\tan ^{ - 1}}\left( {\tan 4} \right)$ , we get $4 - \pi$ as our final answer.
So, the correct answer is “$4 - \pi$”.

Note : Here, to evaluate the given function, we have used the range of inverse tangent which is $- \dfrac{\pi }{2} \leqslant \theta \leqslant \dfrac{\pi }{2}$ and determined the final answer. Similarly, for any trigonometric function, we need to use the range of that particular function to evaluate this type of question. For example, The range of inverse sine is similar to inverse tangent which is $- \dfrac{\pi }{2} \leqslant \theta \leqslant \dfrac{\pi }{2}$ , whereas the range for inverse cosine and inverse cotangent and is $0 \leqslant \theta \leqslant \pi$ . For inverse secant and inverse cosecant, the ranges are $0 \leqslant \theta \leqslant \dfrac{\pi }{2}$ and $- \dfrac{\pi }{2} \leqslant \theta \leqslant 0$ respectively. These ranges play an important role in solving this type of question.