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I am having trouble trying to solve this question:

$\cos \theta \cot\theta(\sec\theta - 2\tan\theta)$

I have had 2 different answers but not the right one.

It's supposed to be: $\cot\theta - 2\cos\theta$

but my skills are rusty and I am having trouble trying to determine exactly what to do. I keep simplifying the identities then multiplying the fractions but I am just super lost..

I am just looking for some help and guidance for this. Thank you.

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    $\begingroup$ $\cos\theta\sec\theta = 1$ and $\cot\theta\tan\theta = 1$... $\endgroup$ – peterwhy Jul 4 '17 at 19:49
  • $\begingroup$ The expression is $ab\left(\dfrac1a-\dfrac2b\right)$. $\endgroup$ – Yves Daoust Jul 4 '17 at 20:43
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$$\cos \theta \cot \theta (\sec \theta -2\tan \theta )=\cos \theta \frac { \cos { \theta } }{ \sin { \theta } } \left( \frac { 1 }{ \cos { \theta } } -\frac { 2\sin { \theta } }{ \cos { \theta } } \right) =\\ =\frac { \cos ^{ 2 }{ \theta } }{ \sin { \theta } } \frac { \left( 1-2\sin { \theta } \right) }{ \cos { \theta } } =\frac { \cos { \theta } }{ \sin { \theta } } \left( 1-2\sin { \theta } \right) =\frac { \cos { \theta } -2\sin { \theta \cos { \theta } } }{ \sin { \theta } } =\frac { \cos { \theta } }{ \sin { \theta } } -\frac { 2\sin { \theta } \cos { \theta } }{ \sin { \theta } } =\cot { \theta } -2\cos { \theta } $$

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  • $\begingroup$ I am confused how $\frac { \cos ^{ 2 }{ \theta } }{ \sin { \theta } } \frac { 1-2\sin { \theta } }{ \cos { \theta } }$ turns into $\frac { \cos { \theta } -2\sin { \theta \cos { \theta } } }{ \sin { \theta } } $ shouldn't it be $\frac { \cos { \theta } -2\sin { \theta } }{ \sin { \theta } }$ since the $\cos^{2}{\theta}$ is divided by the bottom $\cos \theta$ $\endgroup$ – user3670552 Jul 4 '17 at 20:15
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    $\begingroup$ @user3670552 $\cos^2\theta$ when divided by $\cos\theta$ is $\cos\theta$. $\endgroup$ – peterwhy Jul 4 '17 at 20:18
  • $\begingroup$ $cosx$ is reducted in the fraction $\endgroup$ – haqnatural Jul 4 '17 at 20:18
  • $\begingroup$ I know it is reduced when it is divided, but where does the $\cos \theta$ on the right of the $\sin \theta$ come from in the numerator. $\endgroup$ – user3670552 Jul 4 '17 at 20:22
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    $\begingroup$ @user3670552 Do you know that $a(b+c) = ab + ac$? Then $\cos\theta(1-2\sin\theta) = \cos\theta - 2\sin\theta\cos\theta$. $\endgroup$ – peterwhy Jul 4 '17 at 20:26
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You can write $$ \cos\theta\cot\theta(\sec\theta-2\tan\theta)= \cot\theta\cos\theta\sec\theta-2\cos\theta\cot\theta\tan\theta= \cot\theta-2\cos\theta $$ because $\cos\theta\sec\theta=1$ and $\cot\theta\tan\theta=1$.

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$\require {cancel}$ We have $$\cos \theta \cot \theta \ (\sec \theta - 2 \tan \theta)$$ Multiplying everything out (distributive property) we have$$\cot \theta \cos \theta \sec \theta - 2 \cos \theta \cot \theta \tan \theta$$ Note that $\cos \theta \sec \theta = 1$ and $\cot \theta \tan \theta = 1$, so these can cancel... $$\cot \theta \cancelto {1}{\cos \theta \sec \theta} - 2 \cos \theta \cancelto {1}{\cot \theta \tan \theta}$$ And we're left with what we were looking for...$$\cot \theta - 2 \cos \theta$$

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