How do you express #sin^2 theta - sec theta + csc^2 theta # in terms of #cos theta #?
1 Answer
Feb 8, 2016
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
# = 1 - cos^2 theta - 1 / cos theta + 1 / (1 - cos^2 theta)#