How do you prove # tan^2x(1-sin^2x)=(-cos2x+1)/2 #?

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Answer 1 · 2018-07-01T12:37:30

#LHS#

#=\tan^2x(1-\sin^2x)#

#=\frac{\sin^2x}{\cos^2x}(\cos^2x)#

#=\sin^2x#

#=1-\cos^2x#

#1-(\frac{1+\cos2x}{2})#

#=\frac{2-1-\cos2x}{2}#

#=\frac{-\cos 2x+1}{2}#

#=RHS#

Proved.

Answer 2 · 2018-07-01T12:38:12

Please see below.

We know that,

#color(red)((1)sin^2theta+cos^2theta=1#

#color(blue)((2)1-cos2theta=2sin^2theta#

We take ,

#LHS=tan^2x(color(red)(1-sin^2x))tocolor(red)(Apply(1)#

#color(white)(LHS)=sin^2x/cos^2x(color(red)(cos^2x))to[becausetantheta=sintheta/costheta]#

#color(white)(LHS)=sin^2x#

#color(white)(LHS)=1/2[color(blue)(2sin^2x)]tocolor(blue)(Apply(2)#

#color(white)(LHS)=1/2[color(blue)(1-cos2x)]#

#color(white)(LHS)=(-cos2x+1)/2#

#=>LHS=RHS#