TITLE: Why $(\sec^2(x)+1)^{\frac{3}{2}}=\sec^3(x), x>0$?
QUESTION [1 upvotes]: I was trying to find the intermediaries steps to this equality, considering $x$ is a positive number, but I couldn't. Using the identities $\sec^2(x)=1+\tan^2(x)$ and $\sin^2(x)+\cos^2(x)=1$, here some of my work:
First try:
$$\left(\sec^2(x)+1\right)^{\frac{3}{2}}=\left(\dfrac{1}{\cos^2(x)}+1\right)^{\frac{3}{2}}=\left(\dfrac{\cos^2(x)+1}{\cos^2(x)}\right)^{\frac{3}{2}}=\left(\dfrac{(1-\sin^2(x))+1}{\cos^2(x)}\right)^{\frac{3}{2}}$$
Second try:
$$\left(\sec^2(x)+1\right)^{\frac{3}{2}}=\left((\tan^2(x)+1)+1\right)^{\frac{3}{2}}=\left(\tan^2(x)+2\right)^{\frac{3}{2}}$$
Third try (using the first try):
$$\left(\sec^2(x)+1\right)^{\frac{3}{2}}=\left((\cos^2(x)+1)\sec^2(x)\right)^{\frac{3}{2}}=\sec^3(x)\sqrt{(\cos^2(x)+1)^3}$$
In the last and third try, I need that $\sqrt{(\cos^2(x)+1)^3}=1$, then I think the problem is solved...

REPLY [5 votes]: It is not true in general. Indeed, one has:
\begin{align*}
\left(\sec^{2}\left(\dfrac{\pi}{3}\right)+ 1\right)^{3/2}= \dfrac{5\sqrt{5}}{8} \neq \frac{1}{8} = \sec^{3}\left(\dfrac{\pi}{3}\right)
\end{align*}
EDIT
As @JetfiRex has mentioned, you may have switched $\tan^{2}(x)$ by $\sec^{2}(x)$.
If this is the case, the corresponding identity holds:
\begin{align*}
\tan^{2}(x) + 1 & = \frac{\sin^{2}(x)}{\cos^{2}(x)} + 1 = \frac{\sin^{2}(x) + \cos^{2}(x)}{\cos^{2}(x)} = \sec^{2}(x)
\end{align*}
And consequently we also get
\begin{align*}
(\tan^{2}(x) + 1)^{3/2} = (\sec^{2}(x))^{3/2} = |\sec^{3}(x)|
\end{align*}
Now it remains to determine for each values of $x$ one has that $\sec(x) \geq 0$.
Can you take it from here?
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