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

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

• remember $\tan=\dfrac{\sin}{\cos}$ and $sin^2+cos^2=1$ – J. W. Tanner Aug 16 '19 at 16:57

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.

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.