Given #costheta=-4/(sqrt20)#, where #0^@ <= theta <= 180^@# ?

Given #cos(theta)=-4/(sqrt20), 0^@ <= theta <= 180^@#

Use the inverse cosine on both sides:

#theta = cos^-1(-4/(sqrt20)), 0^@ <= theta <= 180^@#

#theta ~~ 153.4^@#

You have the value for the cosine function the denominator should be rationalized:

#cos(theta)= -4/sqrt20#

#cos(theta)= -4/(2sqrt5)#

#cos(theta)= (-2sqrt5)/5#

The secant function is the reciprocal of the cosine function:

#sec(theta) = 1/cos(theta) #

#sec(theta) = -sqrt5/2#

The sine function can be found using the identity:

#sin(theta) = +-sqrt(1-cos^2(theta))#

#sin(theta) = +-sqrt(1-((-2sqrt5)/5)^2)#

#sin(theta) = +-sqrt(25/25- 20/25)#

#sin(theta) = +-sqrt5/5#

We know that the sine function is positive in the second quadrant:

#sin(theta) = sqrt5/5#

The cosecant function is the reciprocal of the sine function:

#csc(theta) = 1/sin(theta)#

#csc(theta) = 5/sqrt5#

#csc(theta) = sqrt5#

Find the tangent function using the identity:

#tan(theta) = sin(theta)/cos(theta)#

#tan(theta) = (sqrt5/5)/((-2sqrt5)/5)#

#tan(theta) = -1/2#

The cotangent function is the reciprocal of the tangent function:

#cot(theta) = -2#