# How do you simplify tanA - cscA secA (1 - 2cos^2 A)?

##### 1 Answer
Nov 13, 2015

Start by putting all of the terms into sine and cosine. This will make it easier to spot identities and see what might cancel out.

$\sin \frac{A}{\cos} A - \frac{1}{\sin} A \frac{1}{\cos} A \left(1 - 2 {\cos}^{2} A\right)$

Multiply through the parenthesis..

$\sin \frac{A}{\cos} A - \frac{1}{\sin A \cos A} + \frac{2 {\cos}^{2} A}{\sin A \cos A}$

Now we should find a common denominator. If we multiply the top and bottom of the tangent term by $\sin A$, we get a denominator of $\sin A \cos A$.

${\sin}^{2} \frac{A}{\sin A \cos A} - \frac{1}{\sin A \cos A} + \frac{2 {\cos}^{2} A}{\sin A \cos A}$

Combine the numerators.

$\frac{{\sin}^{2} A - 1 + 2 {\cos}^{2} A}{\sin A \cos A}$

Now we split the $2 \cos A$ term into $\cos A + \cos A$. I rearranged a little to make the next step more aparent.

$\frac{{\sin}^{2} A + {\cos}^{2} A + {\cos}^{2} A - 1}{\sin A \cos A}$

The Pythagorean theorem states that;

${\sin}^{2} A + {\cos}^{2} A = 1$

Making this substitution into the numerator we get.

$\frac{1 + {\cos}^{2} A - 1}{\sin A \cos A}$

The $1$s cancel out leaving;

${\cos}^{2} \frac{A}{\sin A \cos A}$

The bottom $\cos A$ cancels one from the top, leaving;

$\cos \frac{A}{\sin} A$

This is the identity for;

$\cot A$