How do you simplify #sin^2 B sec^2 B - sec^2 B#?

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Answer 1 · 2018-03-15T23:22:34

-1

Given: #sin^2Bsec^2B-sec^2B#

Apply the reciprocal identity:
#secB=1/cosB#

So: #sin^2B*1/cos^2B-sec^2B#
#sin^2B/cos^2B-sec^2B#

Apply the quotient identity:
#sinB/cosB= TanB#

So: #tan^2B-sec^2B#

Now apply the Pythagorean identity:
#1+tan^2B=sec^2B#
Manipulated to: #-1=tan^2B-sec^2B#

So: #tan^2B-sec^2B= -1#