Question

Find principal solution for $\tan x = - 1$, $x \in \left( {\dfrac{\pi }{2},\pi } \right)$.

Answer 1

We will first consider the given function that is $\tan x = - 1$, $x \in \left( {\dfrac{\pi }{2},\pi } \right)$. As we need to find the principal value, $\arctan$ function is the inverse of $\tan$ function and thus will cancel each other and the required principal value is obtained. While solving the function we will use that $\tan \left( { - x} \right) = - \tan x$ and $\tan \dfrac{\pi }{4} = 1$.

Complete step by step solution: We will first consider the function that is $\tan x = - 1$, $x \in \left( {\dfrac{\pi }{2},\pi } \right)$.
We need to find the principal solution for $\tan x = - 1$.
For finding the principal solution, we must consider that $\arctan$ function is the inverse of $\tan$function and thus will cancel only if $x$ belongs to its principal value.
So, we have, $\tan x = - 1$
We will multiply with $\tan$ inverse on both the sides, we get,

$$\Rightarrow {\tan ^{ - 1}}\left( {\tan x} \right) = {\tan ^{ - 1}}\left( { - 1} \right) \\\ \Rightarrow x = {\tan ^{ - 1}}\left( { - 1} \right) \\\$$

Here, we know that,
$\tan \dfrac{\pi }{4} = 1$ and $\tan \left( { - x} \right) = - \tan x$
Thus, we get,

$$\Rightarrow x = {\tan ^{ - 1}}\left( {\tan \left( { - \dfrac{\pi }{4}} \right)} \right) \\\ \Rightarrow x = {\tan ^{ - 1}}\left( { - \tan \left( {\dfrac{\pi }{4}} \right)} \right) \\\$$

Now, we know that,
${\tan ^{ - 1}}\left( { - \tan x} \right) = \pi - {\tan ^{ - 1}}\left( {\tan x} \right)$
We will substitute the above expression in $x = {\tan ^{ - 1}}\left( { - \tan \left( {\dfrac{\pi }{4}} \right)} \right)$, we get,

$$\Rightarrow x = \pi - {\tan ^{ - 1}}\left( {\tan \left( {\dfrac{\pi }{4}} \right)} \right) \\\ \Rightarrow x = \pi - \dfrac{\pi }{4} \\\ \Rightarrow x = \dfrac{{3\pi }}{4} \\\$$

Thus, we can conclude that the principal solution is $x = \dfrac{{3\pi }}{4}$.

Note: We must remember the trigonometric value of $\tan \dfrac{\pi }{4} = 1$. As we are given the interval in which the principal value lies and we can verify as $x = \dfrac{{3\pi }}{4}$ lies in the interval $x \in \left( {\dfrac{\pi }{2},\pi } \right)$. We know that $\tan$ is negative in second and fourth quadrant so the principal value also lies in these two quadrants. Since, we are asked only in second quadrant so we have find only one principal solution.