Simple and hard trig equation proof

[Guest]

I can't solve this problem [tex]1-4 \cos ( x ) ^ { 2 } +2 \cos ( x ) \sin ( x ) =1- \dfrac{ 4 }{ 1+ \tan ( x ) ^ { 2 } }[/tex]

[Guest]

The first thing I would do is note that [tex]tan(x)= \frac{sin(x)}{cos(x)}[/tex] so we can write the equation as [tex]1- 4cos^2(x)+ 2 cos(x)sin(x)= 1- \frac{4}{1+ \frac{sin^2(x)}{cox^2(x)}}[/tex].

On the right multiply both numerator and denominator by [tex]cos^2(x)[/tex]:
[tex]1- 4cos^2(x)+ 2 cos(x)sin(x)= 1- \frac{4cos^2}{cos^2(x)+ sin^2(x)}=1- 4cos^2(x)[/tex].

Now we have "[tex]1- 4cos^2(x)[/tex]" on both sides which cancel leaving [tex]2cos(x)sin(x)= 1[/tex].

Can you solve that?