Question

Find the principal value of ${\tan ^{ - 1}}\left( { - 1} \right)$.

Answer 1

Hint: The principal value of an inverse trigonometric function at a point x is the value of the inverse function at the point x, which lies in the range of the principal branch. We have to find the principal value of $tan^{-1}(-1)$. We know that the principal value of $tan^{-1}(x)$ is given by $\left( { - \dfrac{{{\pi }}}{2},\dfrac{{{\pi }}}{2}} \right)$ . So, here first we have to find the value of $tan^{-1}(-1)$ using the identity $tan^{-1}(-x) = -(tan^{-1}(x))$. Then ,we can find the principal value.

Complete step-by-step answer:
The values of the tangent functions are-

Function	$0^o$	$30^o$	$45^o$	$60^o$	$90^o$
tan	0	$\dfrac{1}{{\sqrt 3 }}$	1	$\sqrt 3$	Not defined

In the given question we need to find the principal value of ${\tan ^{ - 1}}\left( { - 1} \right)$. We know that for tangent function to be negative, the angle should be negative, that is less than $0^o$.

We know that ${\tan ^{ - 1}}\left( 1 \right) = {45^o}$. This means that the value of $\tan {45^{\text{o}}} = 1$. So the value of the given expression can be calculated as-
$tan^{-1}(-1) = - tan^{-1}(1)$
${\tan ^{ - 1}}\left( { - 1} \right) = - \left( {{{45}^{\text{o}}}} \right) = - {45^{\text{o}}}$
We know that ${{\pi }}$ rad = $180^o$, so
${\tan ^{ - 1}}\left( { - 1} \right) = - \dfrac{{{\pi }}}{4}$
This is the required value.

Note: In such types of questions, we need to strictly follow the range of the principal values that have been specified. The principal value of tangent in the question should always lie between $-90^o$ and $90^o$. This is because there can be infinite values of any inverse trigonometric function. A common mistake is that the students do not remember the formula to eliminate the negative sign. Different functions have different formula according to their principal value ranges, for example-
$cot^{-1}(-x) = 180^{o} - cot^{-1}(x)$
$tan^{-1}(-x) = -(tan^{-1}(x))$