#### Answer

The solution of this system is
$$x=\cos\theta,\quad y=\sin\theta.$$

#### Work Step by Step

Here we treat the sines and cosines as ordinary coefficients. So to solve this problem we will follow the usual steps for solving a system of two equations in two variables:
Step 1: Express $y$ from the second equation:
$$(\cos\theta)y=(\sin\theta)x$$
so we have
$$y=\frac{\sin\theta}{\cos\theta}x = (\tan\theta) x.$$
Step 2: Put this into the first equation and solve for $x$:
$$(\cos\theta)x+x\sin\theta\tan\theta = 1 \Rightarrow x\left(\cos\theta+\sin\theta\tan\theta\right)=1$$
Now remember that $\tan\theta = \sin\theta/\cos\theta$
so for the expression in the bracket on the left we have
$$\cos\theta + \sin\theta\frac{\sin\theta}{\cos\theta} = \cos\theta+ \frac{\sin^2\theta}{\cos\theta} = \frac{\cos^2\theta+\sin^2\theta}{\cos\theta}.$$ Now remember that for any $\theta$ we have that $\sin^2\theta+ \cos^2\theta = 1$ so that the expression in the bracket is finally equal to
$$\frac{1}{\cos\theta}.$$
returning to the equation we now have
$$x\frac{1}{\cos\theta} = 1$$
and this gives
$$x=\cos\theta$$
Step 3: Now use this equality to calculate $y$:
$$y=\frac{\sin\theta}{\cos\theta}x = \frac{\sin\theta}{\cos\theta}\times\cos\theta = \sin\theta.$$