How do you determine #tantheta# given #sectheta=-5/4, 90^circ<theta<180^circ#?

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Answer 1 · 2016-11-28T05:08:33

Please see the explanation.

Here is how you derive a relationship between #tan(theta)# and #sec(theta)#:

Start with #cos^2(theta) + sin^2(theta) = 1#

Divide both sides by #cos^2(theta)#:

#1 + sin^2(theta)/cos^2(theta) = 1/cos^2(theta)#

Substitute #tan^2(theta)# for #sin^2(theta)/cos^2(theta)#

#1 + tan^2(theta) = 1/cos^2(theta)#

Substitute #sec^2(theta)# for #1/cos^2(theta)#:

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

Subtract 1 from both sides:

#tan^2(theta) = sec^2(theta) - 1#

square root both sides:

#tan(theta) = +-sqrt(sec^2(theta) - 1)#

You do not need to remember this, because it is so easily derived.

Now that we have derived the equation for #tan(theta)#, substitute #(-5/4)^2# for #sec^2(theta)#:

#tan(theta) = +-sqrt((-5/4)^2 - 1)#

Because we are told that the angle is in the second quadrant, we know that the tangent function is negative so we drop the + sign:

#tan(theta) = -sqrt((-5/4)^2 - 1)#

#tan(theta) = -sqrt(25/16 - 16/16)#

#tan(theta) = -sqrt(9/16)#

#tan(theta) = -3/4#