## Elementary Linear Algebra 7th Edition

The solution of this system is $$x=\cos\theta,\quad y=\sin\theta.$$
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 $y$: $$y=\frac{\sin\theta}{\cos\theta}x = \frac{\sin\theta}{\cos\theta}\times\cos\theta = \sin\theta.$$