If $\tan (t)+\cot (t)=1$, then one of the values of the expression $\frac{1}{\sin (t)+\cos (t)}$ is:

Given the condition:
$$\tan (t)+\cot (t)=1$$

Convert $\tan (t)$ and $\cot (t)$ into sine and cosine:

$$\frac{\sin (t)}{\cos (t)}+\frac{\cos (t)}{\sin (t)}=1$$

$$\frac{{\sin }^2(t)+{\cos }^2(t)}{\sin (t)\cos (t)}=1$$

Since ${\sin }^2(t)+{\cos }^2(t)=1$:

$$\frac{1}{\sin (t)\cos (t)}=1⟹\sin (t)\cos (t)=1$$

Now, consider the identity for squaring $(\sin (t)+\cos (t))$:

$$(\sin (t)+\cos (t){)}^2={\sin }^2(t)+{\cos }^2(t)+2\sin (t)\cos (t)$$

$$(\sin (t)+\cos (t){)}^2=1+2(1)=3$$

$$\sin (t)+\cos (t)=\pm \sqrt{3}$$

Therefore, the value of the required expression is:

$$\frac{1}{\sin (t)+\cos (t)}=\frac{1}{\pm \sqrt{3}}$$

One of the possible values is $\frac{1}{\sqrt{3}}$.