Between any two real roots of the equation $e^x\sin x-1=0$, the equation $e^x\cos x+1=0$.

Let $f\left(x\right)=e^{-x}-\sin x$ and $\alpha$ and $\beta$ be two roots of the equation $e^x\sin x-1=0$ such that $\alpha <\beta$.

Then  $e^{\alpha }\sin \alpha =1, e^{\beta }\sin \beta =1$

$\Rightarrow e^{-\alpha }-\sin \alpha =0 , e^{-\beta }-\sin \beta =0$

Clearly, $f\left(x\right)$ is continuous on $[\alpha ,\beta ]$ and differentiable on $(\alpha ,\beta )$.

Also $f\left(\alpha \right)=f\left(\beta \right)=0$

$\therefore$ By Rolle's theorem there exists $c\in (\alpha ,\beta )$ such that $f\text{'}\left(c\right)=0$

$$\begin{gathered} \Rightarrow -e^{-c}-\cos c=0 \\ \Rightarrow e^c\cos c+1=0 \end{gathered}$$

So, $x=c$ is root of $e^x\cos x+1=0 , c\in (\alpha ,\beta )$

Hence, option-$1$ is the correct answer.