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

May 5, 2016

The identity you're looking for is ${\csc}^{2} x - 1 = {\cot}^{2} x$.

You can derive this by starting from ${\sin}^{2} x + {\cos}^{2} x = 1$:

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

${\cancel{{\sin}^{2} \frac{x}{\sin} ^ 2 x}}^{1} + \stackrel{{\cot}^{2} x}{\overbrace{{\cos}^{2} \frac{x}{\sin} ^ 2 x}} = \stackrel{{\csc}^{2} x}{\overbrace{\frac{1}{\sin} ^ 2 x}}$

$\setminus m a t h b f \left(1 + {\cot}^{2} x = {\csc}^{2} x\right)$

Thus:

$\textcolor{b l u e}{{\cot}^{2} \frac{x}{\csc x + 1}}$

$= \frac{{\csc}^{2} x - 1}{\csc x + 1}$

$= \frac{\left(\csc x - 1\right) \cancel{\left(\csc x + 1\right)}}{\cancel{\left(\csc x + 1\right)}}$

$= \textcolor{b l u e}{\csc x - 1}$

...if and only if $\csc x + 1 \ne 0$.

If $\csc x + 1 = 0$, then $\frac{1}{\sin} x = - 1$, which is the case when $x = \frac{3 \pi}{2} \pm 2 n \pi$ for all $n \in \mathbb{Z}$.

Therefore, this answer is valid when $x \ne \frac{3 \pi}{2} \pm 2 n \pi$ for all $n \in \mathbb{Z}$.