How do you verify the identity #(1+tan^2theta)/(1-tan^2theta)=1/(2cos^2theta-1)#?

1 Answer
Deepak G.
Aug 7, 2016

#L.H.S=R.H.S.#

Explanation:

Since #sin^2 theta+cos^2 theta=1 #
#L.H.S.=(1+tan^2 theta)/(1-tan^2 theta)#
#=(1+sin^2 theta/cos^2 theta)/(1-sin^2theta/cos^2 theta) #
#=((cos^2 theta+sin^2 theta)/(cancelcos^2 theta))/((cos^2 theta-sin^2 theta)/(cancelcos^2theta)#
#=(cos^2 theta+sin^2 theta)/(cos^2 theta-sin^2 theta)#
#=1/(cos^2 theta-sin^2 theta)#
#=1/(cos^2 theta-(1-cos^2 theta)#
#=1/(cos^2 theta+cos^2 theta-1)#
#=1/(2cos^2 theta-1)#
Hence
#L.H.S=R.H.S.#

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