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Answer 1 · 2017-12-14T06:30:31

See a Proof given in the Explanation.

#"The R.H.S.="(cos2x)(tan2x)(sec^2x),#

#=cancel(cos2x)(sin(2x)/cancelcos(2x))(1/cos^2x),# and, since,

#sin2theta=2sinthetacostheta,# we get,

#"The R.H.S.="(2sinxcosx)(1/cos^2x),#

#=(2sinx)/cosx,#

#2tanx,#

#"=The L.H.S."#

Answer 2 · 2017-12-14T06:34:18

See the proof below

We need

#tanx=sinx/cosx#

#secx=1/cosx#

#sin2x=2sinxcosx#

Therefore,

#RHS=cos(2x)tan(2x)sec^2x#

#=cancel(cos2x)*(sin2x)/(cancel(cos2x))*1/cos^2x#

#=(sin2x)/cos^2x#

#=(2sinxcosx)/cos^2x#

#=2sinx/cosx#

#=2tanx#

#=LHS#

#QED#