Given a right triangle, and its two legs $a$ and $b$, I know of two ways to find the hypotenuse $c$:
Equation 1: $$c=\sqrt{a^2+b^2}$$ Equation 2: $$c=\frac{b}{\sin{(\tan^{-1}{(\frac{b}{a})})}}$$ The first equation is just the pythagorean theorem. The second equation represents finding one of the angles from the legs, and using the definition of $\sin{\theta}=\frac{opp}{hyp}$ to find the hypotenuse. I also know that the expansions of $\sin{x}$ and $\tan^{-1}{x}$ are: $$\sin{x}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}\ldots$$ $$\tan^{-1}{x}=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}\ldots$$ My question is, can one show that the second equation "reduces" to the first equation based on the series expansions of $\sin$ and $\tan^{-1}$ (assuming positive values of $a$ and $b$)? Is this possible?
Are you willing to use the fact that $$ \sin(\tan^{-1} z) = \frac{z}{\sqrt{z^2+1}} ? $$ I can't tell from your question if you would be interested in reducing the second equation to the first by a means other than series. If you only care about series, I can delete this answer.