approximation of law sines from spherical case to planar case we know for plane triangle cosine rule is  $\cos C=\frac{a^+b^2-c^2}{2ab}$ and on spherical triangle is $ \cos C=\frac{\cos c - \cos a  \cos b} {\sin a\sin b}$ suppose $a,b,c<\epsilon$ which are sides of a spherical triangle, and $$|\frac{a^2 +b^2-c^2}{2ab}- \frac{\cos c - \cos a  \cos b} {\sin a\sin b}|<Ke^m$$
could any one tell me what will be $m$ and $K$?

Note that for $x$ close to $0$, 
$$1-\frac{x^2}{2!} \le \cos x\le 1-\frac{x^2}{2!}+\frac{x^4}{4!}$$
and 
$$x-\frac{x^3}{3!} \le \sin x\le x.$$
(We used the Maclaurin series expansion of $\cos x$ and $\sin x$.)
Using these facts on the small angles $a$, $b$, and $c$, we can estimate your difference. 
