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

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Answer 1 · 2016-06-23T05:26:46

#-cos^2x#

#sin^2x=1-cos^2x#

#sec^2x=1/cos^2x#

#tan^2x=(1-cos^2x)/cos^2x#

so, by substituting, you have:

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

#1-cos^2x+(cancel(-1)cancel(+1)-cos^2x)/cos^2x#

#1-cos^2x-1#

#-cos^2x#