If $f(x)=(\frac{1-\tan x}{1+\sin x})^{cosec x}$ is to be made continuous at x = 0, then f(0) should be equal to

$\frac{1}{e^2}$

If f(x) is continuous at x = 0, then $\underset{x→0}{\lim}f(x)=f(0)$

$⇒f(0)=\underset{x→0}{\lim}=e^{\underset{x→0}{\lim}(\frac{-\tan x-\sin x}{(1+\sin x).\sin x})}=e^{\underset{x→0}{\lim}\frac{-(1+\cos x)}{(1+\sin x).\cos x}}=e^{-2}$