1 + tanx.tan2x = sec2x
1+tanx*tan2x = sec 2x
LS =1 +(sin x/cos x)(sin 2x/ cos 2x)
=1 +(sin x/cos x)(2sin x* cos x)/ cos 2x)
=1+2sin^2(x)/(cos 2x)
={cos(2x) +2sin^2(x)}/cos (2x)
= [{cos^2x - sin^2(x) +2sin^2(x)]/cos(2x)
= [{cos^2x +sin^2(x)]/cos(2x)
= [1]/cos(2x)= sec(2x)
Hence proved
LS =1 +(sin x/cos x)(sin 2x/ cos 2x)
=1 +(sin x/cos x)(2sin x* cos x)/ cos 2x)
=1+2sin^2(x)/(cos 2x)
={cos(2x) +2sin^2(x)}/cos (2x)
= [{cos^2x - sin^2(x) +2sin^2(x)]/cos(2x)
= [{cos^2x +sin^2(x)]/cos(2x)
= [1]/cos(2x)= sec(2x)
Hence proved