Given tanθ = - 4/3 and sec θ < 0 , find sinθ and cosθ

Answer 1

For another approach, you can use:

tan^2 (theta) + 1 = sec^2(theta)

(-4/3)^2 +1 = 16/9 +1 = 16/9 + 9/9 = 25/9

So then sec(theta) is plus or minus sqrt(25/9). But since we are told Sec(theta) <0 we know

sec(theta) must be -5/3

Since sec and cos are reciprocals of one another, this gives us

cos (theta) = -3/5

Then sin(theta) / cos(theta) = tan(theta) implies that

sin(theta) = cos(theta)* tan (theta)

= (-3/5)*(-4/3) = 12/15 = 4/5

sin(theta) = 4/5

Answer 2

tangent is negative in quadrants 2 and 4.

cosine (hence the secant) is negative in quadrants 2 and 3.

therefore it must be quadrant 2, where x<0 and y>0

opposite = 4 and adjacent = -3, hypotenuse = 5 by pythagorean

sine = 4/5

cosine = -3/5

tangent = -4/3 as given

cosecant = 5/4

secant = -5/3

cotangent = -3/4