If $\sec \theta =\csc \varphi$ and $0^{\circ }<(\theta ,\varphi )<90^{\circ }$, then the value of $\sin (\theta +\varphi )$ is:

Given the trigonometric relationship:
$$\sec \theta =\csc \varphi$$

We can express this using basic reciprocal definitions:

$$\frac{1}{\cos \theta }=\frac{1}{\sin \varphi }⟹\sin \varphi =\cos \theta$$

Using the complementary angle identity ($\cos \theta =\sin (90^{\circ }-\theta )$):

$$\sin \varphi =\sin (90^{\circ }-\theta )$$

Since both angles are acute ($0^{\circ }<\theta ,\varphi <90^{\circ }$), we can equate their arguments directly:

$$\varphi =90^{\circ }-\theta$$

$$\theta +\varphi =90^{\circ }$$

Now substitute this value to find $\sin (\theta +\varphi )$:

$$\sin (\theta +\varphi )=\sin (90^{\circ })=1$$