How do you identify #(secx-tanx)^2#?

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Answer 1 · 2017-05-04T13:27:09

The answer is #=(1-sinx)/(1+sinx)#

We need

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

#secx=1/cosx#

#tanx=sinx/cosx#

#a^2-b^2=(a+b)(a-b)#

Therefore,

#(secx -tanx)^2#

#=(1/cosx-sinx/cosx)^2#

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

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

#=(1-sinx)^2/((1+sinx)(1-sinx))#

#=(1-sinx)/(1+sinx)#