1. Simplifying using trig identities

Simplify: $\displaystyle \cot(-x) (\cos x) (\tan^2 x + 1)$

I'll show you what I have so far. My problem is I don't really know when I'm finished.

$\displaystyle \cot(-x) (\cos x) (\tan^2 x + 1)$
$\displaystyle (\frac{1}{-\tan x)})(\frac{1}{\sec x})(\sec^2 x)$
$\displaystyle (\frac{1}{-\tan x})(\sec x)$

$\displaystyle (\frac{-\sin x}{\cos x})(\frac{1}{\cos x})$

and I have my final answer as $\displaystyle \frac{-\sin x}{\cos^2 x}$

Is this where I should end it, or should I keep going?

Thanks.

2. Originally Posted by Savior_Self
Simplify: $\displaystyle \cot(-x) (\cos x) (\tan^2 x + 1)$

I'll show you what I have so far. My problem is I don't really know when I'm finished.

$\displaystyle \cot(-x) (\cos x) (\tan^2 x + 1)$
$\displaystyle (\frac{1}{-\tan x)})(\frac{1}{\sec x})(\sec^2 x)$
$\displaystyle (\frac{1}{-\tan x})(\sec x)$

$\displaystyle (\frac{-\sin x}{\cos x})(\frac{1}{\cos x})$ NO!!!

and I have my final answer as $\displaystyle \frac{-\sin x}{\cos^2 x}$

Is this where I should end it, or should I keep going?

Thanks.

$\displaystyle (\frac{1}{-\tan x})(\sec x)$

$\displaystyle = \frac{-1}{\frac{\sin{x}}{\cos{x}}} \sec{x}$

$\displaystyle = \frac{-\cos{x}}{\sin{x}} \frac{1}{\cos{x}}$

$\displaystyle =\frac{-1}{sin{x}}$

$\displaystyle =-\csc{x}$

3. Ahhhhhh, I'm an idiot. Thanks, harish.