How does $\tan^2(x) \sec(x) + \sec^3(x)$ turn in to $2\sec^3(x) - \sec(x)$ Can someone explain how $\tan^2 $ disappeared and $\sec^3$  turn into $2\sec^3$ ???
The derivatives of the function $2\sec(x)\tan(x)$ is apparently $2(-\sec(x) + 2\sec^3(x))$
 A: Need these
$$
\tan=\frac{\sin}{\cos} \tag{$\color{green}{\blacksquare}$}
$$
$$
\sec=\frac{1}{\cos} \tag{$\color{blue}{\blacksquare}$}
$$
$$
\sin^2+\cos^2=1\longrightarrow \sin^2=1-\cos^2 \tag{$\color{red}{\blacksquare}$}
$$
Just use properties and algebra
\begin{align*}
\tan^2\sec+\sec^3
&=
\left(\frac{\sin}{\cos}\right)^2\frac{1}{\cos}+\frac{1}{\cos^3}
\tag*{($\color{blue}{\blacksquare}$),($\color{green}{\blacksquare}$)}\\
&=
\frac{\sin^2}{\cos^2}\frac{1}{\cos}+\frac{1}{\cos^3}
\\&
=
\frac{1}{\cos^3}(\color{red}{\sin^2}+1)
\\&
=
\frac{1}{\cos^3}(\color{red}{1-\cos^2}+1) \tag{$\color{red}{\blacksquare}$}
\\&
=
\frac{1}{\cos^3}({2-\cos^2})
\\&=
2\frac{1}{\cos^3}-\cos^2\frac{1}{\cos^3}\\
&=
2\sec^3-\frac{1}{\cos}
\\&=
\boxed{2\sec^3-\sec} \tag*{$\blacksquare$}
\end{align*}
$\bf IF$ there are any issues ore clarifications needed, $\bf THEN$ comment or send an email to $\tt TheGreatJRB@Berkeley.edu$
Thanks,
Jason
Applied Mathematics Undergraduate
University of California, Berkeley
A: Hint:
$$\sec^2 x - \tan^2 x = 1$$ so $$\tan^2 x = \sec^2 x -1$$
Substitute in your original equation, then multiply.
