Question

Evaluate each of the following:

$\left(\mathrm{i}\right){\tan }^{-1}1+{\cos }^{-1}\left(-\frac{1}{2}\right)+{\sin }^{-1}\left(-\frac{1}{2}\right)$
(ii) ${\tan }^{-1}\left(-\frac{1}{\sqrt{3}}\right)+{\tan }^{-1}\left(-\sqrt{3}\right)+{\tan }^{-1}\left(\sin \left(-\frac{\mathrm{\pi }}{2}\right)\right)$

(iii) ${\tan }^{-1}\left(\tan \frac{5\mathrm{\pi }}{6}\right)+{\cos }^{-1}\left\{\cos \left(\frac{13\mathrm{\pi }}{6}\right)\right\}$

Answer 1

(i) Let ${\sin }^{-1}\left(-\frac{1}{2}\right)=y$
Then,
$\sin y=-\frac{1}{2}$
We know that the range of the principal value branch is $\left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$.
Thus,
$$\begin{gathered} \sin y=-\frac{1}{2}=\sin \left(-\frac{\mathrm{\pi }}{6}\right) \\ \Rightarrow y=-\frac{\mathrm{\pi }}{6}\in \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right] \end{gathered}$$
Now,
Let ${\cos }^{-1}\left(-\frac{1}{2}\right)=z$
Then,
$\cos z=-\frac{1}{2}$
We know that the range of the principal value branch is $\left[0,\mathrm{\pi }\right]$.
Thus,
$$\begin{gathered} \cos z=-\frac{1}{2}=\cos \left(\frac{2\mathrm{\pi }}{3}\right) \\ \Rightarrow z=\frac{2\mathrm{\pi }}{3}\in \left[0,\mathrm{\pi }\right] \end{gathered}$$
So,
${\tan }^{-1}1+{\cos }^{-1}\left(-\frac{1}{2}\right)+{\sin }^{-1}\left(\frac{1}{2}\right)=\frac{\mathrm{\pi }}{4}+\frac{2\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{6}=\frac{3\mathrm{\pi }}{4}$
$\therefore {\tan }^{-1}1+{\cos }^{-1}\left(-\frac{1}{2}\right)+{\sin }^{-1}\left(\frac{1}{2}\right)=\frac{3\mathrm{\pi }}{4}$

(ii)
$$\begin{gathered} {\tan }^{-1}\left(-\frac{1}{\sqrt{3}}\right)+{\tan }^{-1}\left(-\sqrt{3}\right)+{\tan }^{-1}\left(\sin \left(-\frac{\mathrm{\pi }}{2}\right)\right) \\ ={\tan }^{-1}\left(-\frac{1}{\sqrt{3}}\right)+{\tan }^{-1}\left(-\sqrt{3}\right)+{\tan }^{-1}\left(-\sin \left(\frac{\mathrm{\pi }}{2}\right)\right) \\ ={\tan }^{-1}\left(-\frac{1}{\sqrt{3}}\right)+{\tan }^{-1}\left(-\sqrt{3}\right)+{\tan }^{-1}\left(-1\right) \\ =-{\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)-{\tan }^{-1}\left(\sqrt{3}\right)-{\tan }^{-1}\left(1\right) \\ =-{\tan }^{-1}\left(\tan \frac{\mathrm{\pi }}{6}\right)-{\tan }^{-1}\left(\frac{\mathrm{\pi }}{3}\right)-{\tan }^{-1}\left(\frac{\mathrm{\pi }}{4}\right) \\ =-\frac{\mathrm{\pi }}{6}-\frac{\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{4} \\ =-\frac{3\mathrm{\pi }}{4} \end{gathered}$$

(iii)
$$\begin{gathered} {\tan }^{-1}\left(\tan \frac{5\mathrm{\pi }}{6}\right)+{\cos }^{-1}\left\{\cos \left(\frac{13\mathrm{\pi }}{6}\right)\right\} \\ ={\tan }^{-1}\left\{\tan \left(\mathrm{\pi }-\frac{5\mathrm{\pi }}{6}\right)\right\}+{\cos }^{-1}\left\{\cos \left(2\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right)\right\} \\ ={\tan }^{-1}\left\{-\tan \left(\frac{\mathrm{\pi }}{6}\right)\right\}+{\cos }^{-1}\left\{\cos \left(\frac{\mathrm{\pi }}{6}\right)\right\} \\ =-{\tan }^{-1}\left\{\tan \left(\frac{\mathrm{\pi }}{6}\right)\right\}+{\cos }^{-1}\left\{\cos \left(\frac{\mathrm{\pi }}{6}\right)\right\} \\ =-\frac{\mathrm{\pi }}{6}+\frac{\mathrm{\pi }}{6} \\ =0 \end{gathered}$$