Question

How do you evaluate ${{\tan }^{-1}}\left( \tan \left( \dfrac{5\pi }{6} \right) \right)$ ?
(a) Using trigonometric angle identities
(b) Using linear formulas
(c) a and b both
(d) none of the above

Answer 1

In this problem we are trying to evaluate ${{\tan }^{-1}}\left( \tan \left( \dfrac{5\pi }{6} \right) \right)$and we will start by considering $\theta ={{\tan }^{-1}}\left( \tan \left( \dfrac{5\pi }{6} \right) \right)$. And we also know, the value of ${{\tan }^{-1}}$ will be between $-\dfrac{\pi }{2}$ and $\dfrac{\pi }{2}$. Now, using the identity of $\tan a=\tan b\Rightarrow a=n\pi +b$, we get our desired value of $\theta$.

Complete step by step solution:
Let us consider, $\theta ={{\tan }^{-1}}\left( \tan \left( \dfrac{5\pi }{6} \right) \right)$
Now, taking tan function on both sides we get,
$\Rightarrow \tan \theta =\tan \left( \dfrac{5\pi }{6} \right)$
Again we know, from the general solution of $\tan a=\tan b$, we have $\tan a=\tan b\Rightarrow a=n\pi +b$.
So, from the last equation, $\theta =n\pi +\dfrac{5\pi }{6}$ .
Then, if we take, $n=-1$ , we have,
$\theta =-\pi +\dfrac{5\pi }{6}=-\dfrac{\pi }{6}$
And as we all know, the value of ${{\tan }^{-1}}$or arctan function will be between $-\dfrac{\pi }{2}$ and $\dfrac{\pi }{2}$.
Again, if we choose any other number say, n = 0.
Then, $\theta =n\pi +\dfrac{5\pi }{6}$gives us,
$\Rightarrow \theta =0.\pi +\dfrac{5\pi }{6}$
Simplifying, we get,
$\Rightarrow \theta =\dfrac{5\pi }{6}$
This is not taking place between $-\dfrac{\pi }{2}$ and $\dfrac{\pi }{2}$. So, n = 0 is not giving us any solution.
Let us know check another number, say n = 1,
Then, $\theta =n\pi +\dfrac{5\pi }{6}$gives us,
$\Rightarrow \theta =\pi +\dfrac{5\pi }{6}$
Simplifying, we get,
$\Rightarrow \theta =\dfrac{11\pi }{6}$
This is not taking place between $-\dfrac{\pi }{2}$ and $\dfrac{\pi }{2}$. So, n = 1 is not giving us any solution.
Now, from here if we go up on any sides in the positive or negative sides, it will always go out of the range of $-\dfrac{\pi }{2}$ and $\dfrac{\pi }{2}$.
So, only for $n=-1$, this condition satisfies.
Thus,
${{\tan }^{-1}}\left( \tan \left( \dfrac{5\pi }{6} \right) \right)=-\dfrac{\pi }{6}$

So, the correct answer is “Option a”.

Note: A solution of a trigonometric equation is the value of the unknown angle that satisfies the equation. Consider the equation$\sin \theta =\dfrac{1}{2}$ . This equation is, clearly, satisfied by $\theta =\dfrac{\pi }{6},\dfrac{5\pi }{6}$ etc. so these are the solutions of the given equation. Solving an equation means to find the set of all values of the unknown value which satisfy the given equation. The solutions lying between 0 to $2\pi$ or between 0° to 360° are called principal solutions.