Is there an easy way to quickly prove (or memorize) inverse trig formulas such as $ \arcsin(a) = \arctan(\frac{a}{\sqrt{1-a^2}}) $?

Question

Is there an easy way to quickly prove these formulas?

If not, is there any easy mnemonic way to memorize them fast?

$$\begin{align} \arcsin(a) &= \arctan\left(\frac{a}{\sqrt{1-a^2}}\right) \\[4pt] \arccos(a) &= \operatorname{arccot}\left(\frac{a}{\sqrt{1-a^2}}\right) \\[4pt] \arctan(a) = \arcsin\left(\frac{a}{\sqrt{1+a^2}}\right) &= \arccos\left(\frac{1}{\sqrt{1+a^2}}\right) = \operatorname{arccot}\left(\frac{1}{a}\right) \\[4pt] \operatorname{arccot}(a) &= \arccos\left(\frac{a}{\sqrt{1+a^2}}\right) \end{align}$$

P.S. Wikipedia desribes it here

Answer 1

For the first one draw a right angled triangle as below.

Now $\sin x = a \implies x = \sin^{-1}a = \tan^{-1}\frac{a}{\sqrt{1-a^2}}$

Do similarly for the other cases.

Answer 2

For the last one, if $\theta=\operatorname{arccot}(a)$ then $a=\cot\theta,$

so $a^2+1=\left(\dfrac{\cos\theta}{\sin\theta}\right)^2+\left(\dfrac{\sin\theta}{\sin\theta}\right)^2=\dfrac1{\sin^2\theta}=\dfrac1{1-\cos^2\theta}.$

Can you take it from here?

The others are similar.