anonymous
  • anonymous
simplify:-sec^2x/cot^2x+1
Mathematics
katieb
  • katieb
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Hero
  • Hero
\[\space \space \space -\sec^2x \div (\cot^2x + 1)\] \[=-\frac{1}{\cos^2x} \div \left(\frac{\cos^2x}{\sin^2x} + 1\right)\] \[=-\frac{1}{\cos^2x} \div \left(\frac{\cos^2x}{\sin^2x} + \frac{\sin^2x}{\sin^2x}\right)\] \[=-\frac{1}{\cos^2x} \div \left(\frac{1}{\sin^2x} \right)\] \[=-\frac{1}{\cos^2x} \times \left(\frac{\sin^2x}{1} \right)\] \[=-\frac{\sin^2x}{\cos^2x}\] \[=-\tan^2x\]
anonymous
  • anonymous
in my txbook the answer is tan^2x
Hero
  • Hero
There's a negative in front that cannot be ignored

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Hero
  • Hero
Are you sure you wrote down the problem correctly?
anonymous
  • anonymous
k wait
anonymous
  • anonymous
k it was sec not -sec sorry i just wrote it :- instead of writing it :
anonymous
  • anonymous
can u plz help me with other one?
anonymous
  • anonymous
cotx-cos^3xcscx
Hero
  • Hero
\[\cot x - \cos^3x \csc x \\=\frac{\cos x}{\sin x} - \frac{\cos^3x}{\sin x} \\=\frac{\cos x - \cos^3x}{\sin x} \\=\frac{\cos x(1 - \cos^2x)}{\sin x} \\=\frac{\cos x(\sin^2x)}{\sin x} \\=\cos x \sin x \]
anonymous
  • anonymous
thank you so much u rock!i wish u were my pre cal teacher
Hero
  • Hero
okay
anonymous
  • anonymous
wait how did u get cos^3x-cosx=cosx(1-cos^2x) ?
anonymous
  • anonymous
can't we just subtract it out?
anonymous
  • anonymous
k i get what u did !thanks for all ur help

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