anonymous
  • anonymous
help with verifying this please! tanxcos^2x=2tanxcos^2x-tanx/1-tan^2x
Trigonometry
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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chestercat
  • chestercat
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jdoe0001
  • jdoe0001
\(\bf tan(x)cos^2(x) = \cfrac{2tan(x)cos^2(x)-tan(x)}{1-tan^2(x)}\ \ ?\)
anonymous
  • anonymous
yes that's correct
anonymous
  • anonymous
i worked with the right side and converted everything to sin and cos first, but I just confused myself even more

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jdoe0001
  • jdoe0001
$$\bf tan(x)cos^2(x) = \cfrac{2tan(x)cos^2(x)-tan(x)}{1-tan^2(x)}\\ \text{let's do the right-side first}\\ \cfrac{\frac{2sin(x)cos^2(x)}{cos(x)}-\frac{sin(x)}{cos(x)}} {1-\frac{sin^2(x)}{cos^2(x)}}\\ \cfrac{ \frac{2sin(x)cos^2(x)-sin(x)}{cos(x)} }{ \frac{cos^2(x)-sin^2(x)}{cos^2(x)}}\\ \frac{2sin(x)cos^2(x)-sin(x)}{\cancel{cos(x)}} \times \frac{\cancel{cos^2(x)}}{cos^2(x)-sin^2(x)} $$
jdoe0001
  • jdoe0001
$$\bf \frac{sin(x)(2cos^2(x)-1)}{1} \times \frac{cos(x)}{cos^2(x)-sin^2(x)}\\ \cfrac{sin(x)cos(x)(2cos^2(x)-1)}{cos^2(x)-sin^2(x)} $$
jdoe0001
  • jdoe0001
now take a peek at \(\bf 2cos^2(x)-1 \implies cos(2x) \)
jdoe0001
  • jdoe0001
and \(\bf cos(2x) =cos^2(x)-sin^2(x)\) which means that \(\bf 2cos^2(x)-1 = cos^2(x)-sin^2(x)\)
jdoe0001
  • jdoe0001
thus then $$\bf \cfrac{sin(x)cos(x)(2cos^2(x)-1)}{cos^2(x)-sin^2(x)} \implies \cfrac{sin(x)cos(x)\cancel{(cos^2(x)-sin^2(x))}}{\cancel{cos^2(x)-sin^2(x)}} $$
jdoe0001
  • jdoe0001
now try the left side :)

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