$\frac{(1+\mathrm{tg}^2 x)(-1-\sin^2 x)}{2-\cos^2 x}$

Gabriela tried to simplify the expression: $$\frac{(1+\tan^2 x)(-1-\sin^2 x)}{2-\cos^2 x}$$

In which step of her solution did Gabriela make a mistake?

Gabriela´s solution:

First, Gabriela decided to find the domain of the given expression.

• She realized that the denominator $2-\cos^2 x$ is defined $\forall x \in\mathbb{R}$. She observed that $-1\leq\cos^2 x\leq1$, so the denominator does not equal zero for any $x\in\mathbb{R}$.
• Then, she stated that the expression $-1-\sin^2 x$ in the numerator is also defined $\forall x\in\mathbb{R}$.
• Furthermore, she noted that the expression $1+\tan^2 x$ in the numerator is defined $\forall x\in\mathbb{R}\backslash\left\{(2k+1)\frac{\pi}{2};k\in\mathbb{Z}\right\}$.

Gabriela concluded that the domain of the given expression is:

$$\mathbb{R}\backslash\left\{(2k+1)\frac{\pi}{2};k\in\mathbb{Z}\right\}$$ Next, she simplified the given expression for the determined domain ($\forall x\neq(2k+1)\frac{\pi}{2}$, where $k\in\mathbb{Z}$). She performed the following steps. \begin{aligned} \frac{(1+\tan^2 x)(-1-\sin^2 x)}{2-\cos^2 x}&\stackrel{(1)}=\frac{\left(1+\frac{\sin^2 x}{\cos^2 x}\right)\left(-1-\sin^2 x\right)}{2-\cos^2 x}\stackrel{(2)}=\cr &\stackrel{(2)}=\frac{\frac{\cos^2 x+\sin^2 x}{\cos^2 x}\left(-1-\sin^2 x\right)}{2-\cos^2 x} \stackrel{(3)}=\cr &\stackrel{(3)}=\frac{\frac{1}{\cos^2 x}(-\cos^2 x)}{2-\cos^2 x}\stackrel{(4)}=\cr &\stackrel{(4)}=\frac{-1}{2-\cos^2 x} \stackrel{(5)}=\cr &\stackrel{(5)}=\frac{1}{\cos^2 x-2}\end{aligned}