Trigonometry (10th Edition)

$\tan$$^{2} 60^{\circ} + 1 = \sec$$^{2}$ 60$^{\circ}$ First evaluate the value of each trigonometric statement: $\tan$ 60$^{\circ}$ = $\sqrt3$ $\sec$ 60$^{\circ}$ = 2 Next evaluate each side of the equation: $\tan$$^{2} 60^{\circ} + 1 = (\sqrt3)^{2} + 1 = 3 + 1 = 4 \sec$$^{2}$ 60$^{\circ}$ = 2$^{2}$ = 4 Therefore: $\tan$$^{2} 60^{\circ} + 1 = \sec$$^{2}$ 60$^{\circ}$ is a true statement.