Let $\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}-2\mathrm{sinx}-\mathrm{\pi }\le \mathrm{x}\le -\frac{\mathrm{\pi }}{2}\\ \mathrm{asinx}+\mathrm{b} -\frac{\mathrm{\pi }}{2}<\mathrm{x}<\frac{\mathrm{\pi }}{2}\\ \mathrm{cosx} \frac{\mathrm{\pi }}{2}\le \mathrm{x}\le \mathrm{\pi }\end{array}\right.$ If $f$ is continuous on $[-\pi ,\pi )$, then 

The given function is $\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}-2\mathrm{sinx}-\mathrm{\pi }\le \mathrm{x}\le -\frac{\mathrm{\pi }}{2}\\ \mathrm{asinx}+\mathrm{b} -\frac{\mathrm{\pi }}{2}<\mathrm{x}<\frac{\mathrm{\pi }}{2}\\ \mathrm{cosx} \frac{\mathrm{\pi }}{2}\le \mathrm{x}\le \mathrm{\pi }\end{array}\right.$

Since $f\left(x\right)$ s continuous on $(-\pi ,\pi )$

$\Rightarrow f$ is continuous at $x=-\frac{\pi }{2}$ and at $x=\frac{\pi }{2}$

$\Rightarrow \underset{x\to -{\frac{\pi }{2}}^{-}}{\lim }f\left(x\right)=\underset{x\to -{\frac{\pi }{2}}^{+}}{\lim }f\left(x\right)$

$\Rightarrow -2\sin \left(\frac{-\pi }{2}\right)=a\sin \left(\frac{-\mathrm{\pi }}{2}\right)+b$

$\Rightarrow 2=\mathrm{b}-\mathrm{a}...\left(1\right)$

And $\underset{x\to {\frac{\pi }{2}}^{-}}{\lim }f\left(x\right)=\underset{x\to {\frac{\pi }{2}}^{+}}{\lim }f\left(x\right)$

$$\begin{gathered} \Rightarrow \mathrm{asin}\left(\frac{\mathrm{\pi }}{2}\right)+\mathrm{b}=\cos \left(\frac{\mathrm{\pi }}{2}\right) \\ \Rightarrow \mathrm{a}+\mathrm{b}=0. \\ \Rightarrow \mathrm{a}=-\mathrm{b}...\left(2\right) \end{gathered}$$

From equations $\left(1\right), \left(2\right)$ we get, $\mathrm{b}=1, \mathrm{a}=-1$

Hence option-$1$ is the correct answer.