I'm aware of how we can simplify functions which have $Arc$ as an argument . For example $\sin(\cos^{-1}(x)) = \sqrt{1-x^2}$ but what about cases which $Arc$ is out of the parentheses ? For instance consider this : $\sin^{-1}(\tan x)$ . Is there any way for simplification ?
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$\begingroup$You can squeeze some more out of this.
Consider $$ f = \ln ( \sec x + \tan x)$$
It’s easy to see that $f’ = \sec x$. Now consider $g = sin^{-1} (i \tan(x))$
We have that $g’ = \sec(x)$
So there is some constant $K$ such that:
$$ \ln ( \sec x + \tan x) + K = \sin^{-1} (i \tan(x))$$
$$ \ln ( ( (\frac{i \tan x }{i}) ^2 + 1) ^{1/2}+ \frac{i \tan x }{i} )+ K = sin^{-1} (i \tan(x))$$
So we can conclude:
$$ \ln ( \sqrt{ 1 - \tan^2} - i \tan(x) ) = \sin^{-1} \tan x $$
Which might be a little more insightful.
$\endgroup$ $\begingroup$Well, let $f(x)=\arcsin(\tan(x))$, $x\in(-\frac{\pi}{4},\frac{\pi}{4})$. Now, since: $$\arcsin(x)=\int_0^{x}\frac{1}{\sqrt{1-t^2}}dt$$ we have, that: $$f(x)=\int_0^{\tan(x)}\frac{1}{\sqrt{1-t^2}}dt$$ So: $$\begin{align*}f'(x)=&\frac{1}{\sqrt{1-\tan^2(x)}}\frac{1}{\cos^2(x)}=\frac{1}{\cos^2(x)\sqrt{1-\frac{\sin^2(x)}{\cos^2(x)}}}=\\=&\frac{1}{\cos^2(x)\sqrt{\frac{cos^2(x)-\sin^2(x)}{\cos^2(x)}}}=\frac{1}{\cos(x)\sqrt{\cos(2x)}}\end{align*}$$ So: $$f(x)=\int_0^x\frac{1}{\cos(t)\sqrt{\cos(2t)}}dt+c$$ and since $f(0)=\arcsin(\tan(0))=0$, we have $c=0$, so: $$f(x)=\int_0^x\frac{1}{\cos(t)\sqrt{\cos(2t)}}d,\ x\in\left(-\frac{\pi}{4},\frac{\pi}{4}\right)$$
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