## Trigonometry (11th Edition) Clone

(Without a calculator) $60^{o}$ is a Special Angle, so from the table of Function Values of Special Angles, we find $\tan 60^{o}=\sqrt{3},\qquad \sec 60^{o}=2$ $LHS=1+(\sqrt{3})^{2}=1+3=4$ $RHS=(2)^{2}=4$ $LHS=RHS$ the statement is true