(e^x − cos x) over sin x

Idea. Divide top and bottom by $x$ to make familiar quotients appear: for $x \neq 0$,
$$\frac{e^x-\cos x}{\sin x} = \frac{\frac{e^x-\cos x}{x}}{\frac{\sin x}{x}}.$$

Numerator. For $x\neq0$,
$$\frac{e^x-\cos x}{x}=\frac{e^x-1}{x}+\frac{1-\cos x}{x}.$$
The first quotient tends to $1$. For the second, write $$\dfrac{1-\cos x}{x}=x\cdot\dfrac{1-\cos x}{x^2}\to0\cdot\dfrac12=0.$$ So the numerator divided by $x$ tends to $1+0=1$.

Denominator. $\dfrac{\sin x}{x}\to1$.

Conclusion.
$$\frac{e^x-\cos x}{\sin x}=\frac{(e^x-\cos x)/x}{(\sin x)/x}\longrightarrow\frac11=1.$$