Answer
true
Work Step by Step
(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
