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 ,

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

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

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

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

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

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

$=>LHS=RHS$

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