How does one verify #(1-sec^2x)/sec^2x=-sin^2x#?

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#(1-sec^2x)/sec^2x=-sin^2x#

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Answer 1 · 2018-04-23T00:48:02

# frac{ 1- sec^2 x}{sec^2 x} ##= frac { 1 - 1/cos^2 x}{1/cos^2 x} ##= cos ^2 x ( 1 - 1/cos^2 x ) ## = \cos ^2 x - 1 ## = - sin^2 x #

# frac{ 1- sec^2 x}{sec^2 x} #

#= frac { 1 - 1/cos^2 x}{1/cos^2 x} #

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

# = \cos ^2 x - 1 #

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

# = - sin^2 x #