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 $2(-\sec(x) + 2\sec^3(x))$

Solution 1:

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*}

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Thanks,
Jason
Applied Mathematics Undergraduate
University of California, Berkeley