Answer
$\sin$(-510) = -$\frac{1}{2}$
$\cos$(-510) = -$\frac{\sqrt3}{2}$
$\tan$(-510) = $\frac{\sqrt3}{3}$
$\cot$(-510) = $\sqrt3$
$\csc$(-510) = -2
$\sec$(-510) = -$\frac{2\sqrt3}{3}$
Work Step by Step
-510$^{\circ}$
We must first fine the coterminal angle:
-510$^{\circ}$ + 360$^{\circ}$ = -150$^{\circ}$
-150 is in Quadrant III. Therefore all the trigonometric functions are negative with the exception of $\tan$ and $\cot$
The reference angle is:
$\theta$$^{1}$ = -150$^{\circ}$ +`180$^{\circ}$ = 30$^{\circ}$
$\sin$(30) = -$\frac{1}{2}$
$\cos$(30) = -$\frac{\sqrt3}{2}$
$\tan$(30) = $\frac{\sqrt3}{3}$
$\cot$(30) = $\sqrt3$
$\csc$(30) = -2
$\sec$(30) = -$\frac{2\sqrt3}{3}$
