Question

help with verifying this please!
tanxcos^2x=2tanxcos^2x-tanx/1-tan^2x

Answer 1

\(\bf tan(x)cos^2(x) = \cfrac{2tan(x)cos^2(x)-tan(x)}{1-tan^2(x)}\ \ ?\)

Answer 2

yes that's correct

Answer 3

i worked with the right side and converted everything to sin and cos first, but I just confused myself even more

Answer 4

$$\bf
tan(x)cos^2(x) = \cfrac{2tan(x)cos^2(x)-tan(x)}{1-tan^2(x)}\\
\text{let's do the right-side first}\\
\cfrac{\frac{2sin(x)cos^2(x)}{cos(x)}-\frac{sin(x)}{cos(x)}}
{1-\frac{sin^2(x)}{cos^2(x)}}\\
\cfrac{
\frac{2sin(x)cos^2(x)-sin(x)}{cos(x)}
}{
\frac{cos^2(x)-sin^2(x)}{cos^2(x)}}\\
\frac{2sin(x)cos^2(x)-sin(x)}{\cancel{cos(x)}} \times \frac{\cancel{cos^2(x)}}{cos^2(x)-sin^2(x)}
$$

Answer 5

$$\bf \frac{sin(x)(2cos^2(x)-1)}{1} \times \frac{cos(x)}{cos^2(x)-sin^2(x)}\\
\cfrac{sin(x)cos(x)(2cos^2(x)-1)}{cos^2(x)-sin^2(x)}
$$

Answer 6

now take a peek at \(\bf 2cos^2(x)-1 \implies cos(2x) \)

Answer 7

and \(\bf cos(2x) =cos^2(x)-sin^2(x)\)

which means that

\(\bf 2cos^2(x)-1 = cos^2(x)-sin^2(x)\)

Answer 8

thus then
$$\bf
\cfrac{sin(x)cos(x)(2cos^2(x)-1)}{cos^2(x)-sin^2(x)}
\implies \cfrac{sin(x)cos(x)\cancel{(cos^2(x)-sin^2(x))}}{\cancel{cos^2(x)-sin^2(x)}}
$$

Answer 9

now try the left side :)