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ylucho
help with verifying this please! tanxcos^2x=2tanxcos^2x-tanx/1-tan^2x
\(\bf tan(x)cos^2(x) = \cfrac{2tan(x)cos^2(x)-tan(x)}{1-tan^2(x)}\ \ ?\)
i worked with the right side and converted everything to sin and cos first, but I just confused myself even more
$$\bf tan(x)cos^2(x) = \cfrac{2tan(x)cos^2(x)-tan(x)}{1-tan^2(x)}\\ \text{let's do the right-side first}\\ \cfrac{\frac{2sin(x)cos^2(x)}{cos(x)}-\frac{sin(x)}{cos(x)}} {1-\frac{sin^2(x)}{cos^2(x)}}\\ \cfrac{ \frac{2sin(x)cos^2(x)-sin(x)}{cos(x)} }{ \frac{cos^2(x)-sin^2(x)}{cos^2(x)}}\\ \frac{2sin(x)cos^2(x)-sin(x)}{\cancel{cos(x)}} \times \frac{\cancel{cos^2(x)}}{cos^2(x)-sin^2(x)} $$
$$\bf \frac{sin(x)(2cos^2(x)-1)}{1} \times \frac{cos(x)}{cos^2(x)-sin^2(x)}\\ \cfrac{sin(x)cos(x)(2cos^2(x)-1)}{cos^2(x)-sin^2(x)} $$
now take a peek at \(\bf 2cos^2(x)-1 \implies cos(2x) \)
and \(\bf cos(2x) =cos^2(x)-sin^2(x)\) which means that \(\bf 2cos^2(x)-1 = cos^2(x)-sin^2(x)\)
thus then $$\bf \cfrac{sin(x)cos(x)(2cos^2(x)-1)}{cos^2(x)-sin^2(x)} \implies \cfrac{sin(x)cos(x)\cancel{(cos^2(x)-sin^2(x))}}{\cancel{cos^2(x)-sin^2(x)}} $$
now try the left side :)