How do you simplify tan^2x - (cot^2x + 1)/cot^2x?

Dec 4, 2016

Use the identity $\tan x = \sin \frac{x}{\cos} x$ and $\cot x = \frac{1}{\tan} x = \frac{1}{\sin \frac{x}{\cos} x} = \cos \frac{x}{\sin} x$.

So ${\tan}^{2} x - \frac{{\cot}^{2} x + 1}{\cot} ^ 2 x$

$= {\sin}^{2} \frac{x}{\cos} ^ 2 x - \left(\frac{{\cos}^{2} \frac{x}{\sin} ^ 2 x + 1}{{\cos}^{2} \frac{x}{\sin} {x}^{2}}\right)$

$= {\sin}^{2} \frac{x}{\cos} ^ 2 x - \frac{\frac{{\cos}^{2} x + {\sin}^{2} x}{\sin} ^ 2 x}{{\cos}^{2} \frac{x}{\sin} ^ 2 x}$

Now use ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$.

$= {\sin}^{2} \frac{x}{\cos} ^ 2 x - \left(\frac{1}{\sin} ^ 2 x \times {\sin}^{2} \frac{x}{\cos} ^ 2 x\right)$

$= {\sin}^{2} \frac{x}{\cos} ^ 2 x - \frac{1}{\cos} ^ 2 x$

$= \frac{{\sin}^{2} x - 1}{\cos} ^ 2 x$

$= \frac{- {\cos}^{2} x}{\cos} ^ 2 x$

$= - 1$

Hopefully this helps!

Dec 4, 2016

We can make use of the Pythagorean identities:
${\sin}^{2} x + {\cos}^{2} x = 1$
$\text{ "1" } + {\cot}^{2} x = {\csc}^{2} x$
${\tan}^{2} x + \text{ "1" } = {\sec}^{2} x$

Explanation:

${\tan}^{2} x - \frac{\textcolor{g r e e n}{{\cot}^{2} x + 1}}{\textcolor{n a v y}{{\cot}^{2} x}} = {\tan}^{2} x - \frac{\textcolor{g r e e n}{{\csc}^{2} x}}{\textcolor{n a v y}{{\cot}^{2} x}}$
$\textcolor{w h i t e}{{\tan}^{2} x - \frac{{\cot}^{2} x + 1}{\cot} ^ 2 x} = {\tan}^{2} x - \frac{\textcolor{g r e e n}{1 / \cancel{{\sin}^{2} x}}}{\textcolor{n a v y}{{\cos}^{2} x / \cancel{{\sin}^{2} x}}}$
$\textcolor{w h i t e}{{\tan}^{2} x - \frac{{\cot}^{2} x + 1}{\cot} ^ 2 x} = {\tan}^{2} x - {\sec}^{2} x$
$\textcolor{w h i t e}{{\tan}^{2} x - \frac{{\cot}^{2} x + 1}{\cot} ^ 2 x} = - 1$

You can also convert all the trig functions to $\sin$'s and $\cos$'s if you are not sure what to do; it's sort of a catch-all. It will always work, but it may take a bit longer.