00:02
In this problem, we're given the information, tangent of x equals square root 3 over 3, cosine of x equals negative square root 3 over 2, and we want to find the values of the rest of the trig functions.
00:16
So it's going to work out more easily if we change tangent to its equivalent unrationalized form, which would be 1 over square root 3.
00:27
And cotangent is the reciprocal of tangent.
00:30
So co -tangent x would be square root 3 over 1, which we're just going to write a square root 3.
00:37
Secant is the reciprocal of cosine.
00:40
So then the reciprocal of negative 3 over, excuse me, of negative square root 3 over 2 is negative 2 over square root 3.
00:48
And we would rationalize that by multiplying the top and bottom by square root 3, and we end up with negative 2 square root 3 over 3.
00:58
And then to find sign in co -secant, i'm going to use a reference triangle.
01:02
Notice that tangent is positive and cosine is negative, so that means we're going to be in quadrant 3...