1. ## Inverse Trigonometry

Hello all,
just after a bit of help with this question, do i show this by taking the derivative? i'm not sure..

Show that
$\tan^{-1}x=\sin^{-1}\frac{x}{\sqrt {1+x^2}}$ for all $x \in (-\infty, \infty)$

2. $\tan v = \frac{\sin v}{ \cos v}=\frac{x}{1} \implies \arctan \frac{x}{1} = \arcsin \frac{x}{\sqrt{1+x^2}}$

As you can see above, $\sin v = x$ and $\cos v = 1$. The hypotenuse then becomes $\sqrt{1+x^2}$ (Pythagoras' theorem).

3. Hello, Robb!

No derivatives needed . . . It's an Identity problem.

Show that: . $\tan^{\text{-}1}x \;=\;\sin^{\text{-}1}\!\left(\frac{x}{\sqrt {1+x^2}}\right)$ . for all $x \in (\text{-}\infty, \infty)$
$\text{Let: }\:\theta \:=\:\tan^{\text{-}1}x$ .[1]

Then we have: . $\tan\theta \:=\:\frac{x}{1} \:=\:\frac{opp}{adj}$

$\theta$ is in a right triangle with: $opp = x,\;adj = 1$

Using Pythagorus, we have: . $hyp = \sqrt{1+x^2}$

Hence: . $\sin\theta \:=\:\frac{opp}{hyp} \:=\:\frac{x}{\sqrt{1+x^2}} \quad\Rightarrow\quad \theta \:=\:\sin^{\text{-}1}\!\left(\frac{x}{\sqrt{1+x^2}}\right)$ .[2]

Equate [1] and [2]: . $\tan^{\text{-}1}x \;=\;\sin^{\text{-}1}\!\left(\frac{x}{\sqrt{1+x^2}}\right)$