Prove that#( cot^2A+1)/(cot^2A-1)=sec2A#?

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Answer 1 · 2017-11-14T10:55:08

We need

#cotA=cosA/sinA#

#cos^2A-sin^2A=cos2A#

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

Therefore,

#LHS=(cot^2A+1)/(cot^2A-1)#

#=(cos^2A/sin^2A+1)/(cos^2A/sin^2A-1)#

#=(cos^2A+sin^2A)/(cos^2A-sin^2A)#

#=1/cos(2A)#

#=sec(2A)#

#=RHS#

#QED#

Answer 2 · 2017-11-14T17:44:22

Refer to a Proof of in the Explanation.

We know that, #cos2A=(1-tan^2A)/(1+tan^2A).#

#:. sec2A=1/(cos2A),#

#=1/{(1-tan^2A)/(1+tan^2A),#

#=(1+tan^2A)/(1-tan^2A),#

Replacing #tanA# by #1/cotA,# we have,

#sec2A={1+1/cot^2A)/(1-1/cot^2A),#

#=(cot^2A+1)/(cot^2A-1).#

Hence, the Proof.