Trouble simplifying this Trigonometric Identity

I am having trouble trying to solve this question:

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

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.

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

$$\cos \theta \cot \theta (\sec \theta -2\tan \theta )=\cos \theta \frac { \cos { \theta } }{ \sin { \theta } } \left( \frac { 1 }{ \cos { \theta } } -\frac { 2\sin { \theta } }{ \cos { \theta } } \right) =\\ =\frac { \cos ^{ 2 }{ \theta } }{ \sin { \theta } } \frac { \left( 1-2\sin { \theta } \right) }{ \cos { \theta } } =\frac { \cos { \theta } }{ \sin { \theta } } \left( 1-2\sin { \theta } \right) =\frac { \cos { \theta } -2\sin { \theta \cos { \theta } } }{ \sin { \theta } } =\frac { \cos { \theta } }{ \sin { \theta } } -\frac { 2\sin { \theta } \cos { \theta } }{ \sin { \theta } } =\cot { \theta } -2\cos { \theta }$$
• 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 $\cos \theta$ – user3670552 Jul 4 '17 at 20:15
• @user3670552 $\cos^2\theta$ when divided by $\cos\theta$ is $\cos\theta$. – peterwhy Jul 4 '17 at 20:18
• $cosx$ is reducted in the fraction – haqnatural Jul 4 '17 at 20:18
• 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. – user3670552 Jul 4 '17 at 20:22
• @user3670552 Do you know that $a(b+c) = ab + ac$? Then $\cos\theta(1-2\sin\theta) = \cos\theta - 2\sin\theta\cos\theta$. – peterwhy Jul 4 '17 at 20:26
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 $\cot\theta\tan\theta=1$.
$\require {cancel}$ 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...$$\cot \theta - 2 \cos \theta$$