Question #2d72f

1 Answer
Narad T.
Dec 18, 2017

See the proof below

Explanation:

Reminder :

#1+tan^2x=sec^2x#

#cos2x=cos^2x-sin^2x#

#sin2x=2sinxcosx#

Therefore,

#LHS=(cos2x)/(1-sin2x)#

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

Dividing by #cos^2x#

#LHS=(1-(sin^2x/cos^2x))/(1/cos^2x-(2sinxcosx)/(cos^2x))#

#=(1-tan^2x)/(sec^2x-2tanx)#

#=(1-tan^2x)/(1+tan^2x-2tanx)#

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

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

#=RHS#

#QED#

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