Question Number 16788 by jyoti5726 last updated on 26/Jun/17
Commented by 1234Hello last updated on 04/Jul/17
$$\mathrm{tan}^{−\mathrm{1}} \:\sqrt{\mathrm{3}}\:=\:\frac{\pi}{\mathrm{3}} \\ $$$$\mathrm{sec}^{−\mathrm{1}} \:\left(−\mathrm{2}\right)\:=\:\mathrm{cos}^{−\mathrm{1}} \:\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)\:=\:\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\mathrm{tan}^{−\mathrm{1}} \:\sqrt{\mathrm{3}}\:−\:\mathrm{sec}^{−\mathrm{1}} \:\left(−\mathrm{2}\right)\:=\:\frac{−\pi}{\mathrm{3}} \\ $$