If $x \cos\theta+y\sin\theta=a$ and $x\sin\theta-y\cos\theta=b$, then $\tan\theta=\frac{bx+ay}{ax-by}$. (Math Olympiad)

Question

I tried to solve it but I can’t get the answer. Please help me in proving this trig identity:

If $$x \cos\theta+y\sin\theta=a$$ $$x\sin\theta-y\cos\theta=b$$ then $$\tan\theta=\frac{bx+ay}{ax-by}$$

I've spent many hours trying. Thanks in advance.

Answer 1

For fun, here's a trigonograph:

$$\tan\theta = \tan(\phi + \psi) = \frac{\tan\phi+\tan\psi}{1-\tan\phi\tan\psi} = \frac{\;\dfrac{y}{x}+\dfrac{b}{a}\;}{\;1-\dfrac{y}{x}\dfrac{b}{a}\;}=\frac{ay+bx}{ax-by}$$

Answer 2

Hint:

Solve the two simultaneous linear equation for $\sin\theta,\cos\theta$

See https://brilliant.org/wiki/system-of-linear-equations/ or https://revisionmaths.com/gcse-maths-revision/algebra/simultaneous-equations

Answer 3

We first try to find $\sin \theta $ in terms of $a, b, x, y$

Multiply equation first by $y$ and the equation $2^{nd}$ by $x$ and add the resultant equations to get $$\sin \theta= \frac {bx+ay}{x^2+y^2}$$

Now multiply the first equation by $x$ and second equation by $y$ to get following equations

$$x^2\cos \theta+xy\sin \theta=ax$$ And $$xy\sin\theta-y^2\cos\theta=by$$

Hence we get $$\cos\theta=\frac {ax-by}{x^2+y^2}$$ By subtracting resultant $2^{nd}$ equation from the resultant $1^{st}$ equation.

Using these values of $\sin\theta$ and $\cos\theta$ we get $$\tan\theta = \frac {bx+ay}{ax-by}$$