How do you express #sin^2 theta - sec theta + csc^2 theta # in terms of #cos theta #?

1 Answer
Lotusbluete
Feb 8, 2016

#1 - cos^2 theta - 1 / cos theta + 1 / (1 - cos^2 theta)#

Explanation:

You should use the following identities:

[1] #" "sin^2 theta + cos^2 theta = 1 " "<=>" " sin^2 theta = 1 - cos^2 theta#

[2] #" "sec theta = 1 / cos theta#

[3] #" "csc theta = 1 / sin theta#

Thus, your expression can be transformed as follows:

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

# = 1 - cos^2 theta - 1 / cos theta + 1 / sin^2 theta#

.... use #sin^2 theta = 1 - cos^2 theta# once again...

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

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