If $\int \frac{dx}{1+\sin x}=\tan \left(\frac{{\displaystyle x}}{{\displaystyle 2}}+a\right)+b$ then a =

$\int \frac{1}{1+\sin x}\times \frac{1-\sin x}{1-\sin x}dx=\int \frac{1-\sin x}{1-{\sin }^{2}x}dx=\int \frac{1-\sin x}{{\cos }^{2}x}dx$

$=$$\int \left(\frac{1}{{\cos }^{2}x}-\frac{\sin x}{{\cos }^{2}x}\right)dx=\int {\sec }^{2}xdx-\int \tan x.\sec xdx$

$=tanx-secx+c$

$=\frac{\sin x}{\cos x}-\frac{1}{\cos x}+c$

$=-\frac{(1-\sin x)}{\cos x}+c$

$=-\frac{{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}^2}{{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}}+c$

$=-\frac{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}{\cos \frac{x}{2}+\sin \frac{x}{2}}+c$

$=-\left(\frac{{\displaystyle 1-\tan \frac{{\displaystyle x}}{{\displaystyle 2}}}}{{\displaystyle 1+\tan \frac{{\displaystyle x}}{{\displaystyle 2}}}}\right)+c$

$=-\tan \left(\frac{{\displaystyle \pi }}{{\displaystyle 4}}-\frac{{\displaystyle x}}{{\displaystyle 2}}\right)+c$

$=\tan \left(\frac{{\displaystyle x}}{{\displaystyle 2}}-\frac{{\displaystyle \pi }}{{\displaystyle 4}}\right)+c$

$\therefore a=\frac{-\pi }{4}$