# Question #b4e98

Apr 15, 2015

Let's start from a definition of the functions mentioned in the problem.

(1) Evaluation of ${\tan}^{-} 1 \left(0\right)$
By definition, $\tan \left(\phi\right) = \sin \frac{\phi}{\cos} \left(\phi\right)$.
Therefore, ${\tan}^{-} 1 \left(\phi\right) = \cos \frac{\phi}{\sin} \left(\phi\right)$
By definition, $\cos \left(\phi\right)$ is an abscissa of a point on a unit circle that is an endpoint of a radius at angle $\phi$ to the X-axis. Also by definition, $\sin \left(\phi\right)$ is an ordinate of this point.

If an angle $\phi = 0$, the abscissa $\cos \left(0\right)$ of an endpoint of a corresponding radius equals to $1$, while it's ordinate $\sin \left(0\right)$ equals to $0$.
As we see, the denominator of the expression for
${\tan}^{-} 1 \left(0\right) = \cos \frac{0}{\sin} \left(0\right)$ equals to $0$, which means that ${\tan}^{-} 1 \left(0\right)$ is UNDEFINED.

(2) Evaluation of ${\csc}^{-} 1 \left(2\right)$
By definition, $\csc \left(\phi\right) = \frac{1}{\sin} \left(\phi\right)$
Therefore, ${\csc}^{-} 1 \left(\phi\right) = \sin \left(\phi\right)$
As we stated above, by definition, $\sin \left(\phi\right)$ is an ordinate of the endpoint of a radius that forms an angle $\phi$ with the X-axis.

If an angle $\phi = 2$ (that is, $2$ radians), the abscissa of the endpoint of a corresponding radius is, approximately, $0.91$. That is the value of ${\csc}^{-} 1 \left(2\right)$.