Taking L.H.S = \[2\text{ }ta{{n}^{-1}}\left( -3 \right)\text{ }=\text{ }-2\text{ }ta{{n}^{-1}}~3\text{ }(\because ta{{n}^{-1}}~\left( -x \right)\text{ }=\text{ }\text{ }ta{{n}^{-1}}~x)\text{ }\in R)\]
= R.H.S
– Hence Proved.
Show that \[\mathbf{2}\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}\left( -\mathbf{3} \right)\text{ }=\text{ }\text{ }\mathbf{\pi }/\mathbf{2}\text{ }+\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}\left( -\mathbf{4}/\mathbf{3} \right)\]
Show that \[\mathbf{2}\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}\left( -\mathbf{3} \right)\text{ }=\text{ }\text{ }\mathbf{\pi }/\mathbf{2}\text{ }+\text{ }\mathbf{ta}{{\mathbf{n}}^{-\mathbf{1}}}\left( -\mathbf{4}/\mathbf{3} \right)\]