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 $$\tan\theta = \frac {bx+ay}{ax-by}$$

Answer 4

$$x\cos \theta+y\sin\theta=a$$

$$\implies x+y\tan\theta=a\sec\theta$$

$$\implies \sec \theta=\frac{x+y\tan\theta}{a} ...(1)$$

$$x\sin\theta-y\cos\theta=b$$

$$\implies x\tan\theta-y=b\sec\theta$$

$$\implies \sec\theta=\frac{x\tan\theta-y}{b}...(2)$$ From $1$ and $2$,

$$\frac{x+y\tan\theta}{a}=\frac{x\tan\theta-y}{b}$$

$$\implies bx+by\tan\theta=ax\tan\theta-ay$$

$$\implies \tan\theta(ax-by)=bx+ay$$

$$\tan\theta=\frac{bx+ay}{ax-by}$$

Answer 5

$x \cos\theta + y\sin\theta = a, \,\, x \sin\theta - y\cos\theta = b$. Therefore:

$$\dfrac{x \cos\theta + y\sin\theta}{x \sin\theta - y\cos\theta} = \dfrac{a}{b}$$ $$\implies \dfrac{x + y\tan\theta}{x \tan\theta - y} = \dfrac{a}{b}$$ $$\implies bx + by\tan\theta = ax \tan\theta - ay$$ $$\implies (ax - by) \tan\theta = bx + ay$$ $$\implies \tan\theta = \dfrac{bx + ay}{ax - by}$$

Answer 6

We may write the given equations as: $$\begin{pmatrix}\cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix}x \\ - y \end{pmatrix} = \begin{pmatrix}a \\ b \end{pmatrix} $$ and recognise that the first matrix represents a counterclockwise rotation by $\theta$ about the origin. We can then use the formula for the tangent of the difference of two angles to see that $$\tan \theta = \frac{\frac{b}{a}- \frac{-y}{x}}{1+\frac{b}{a} \frac{-y}{x}}=\frac{bx+ay}{ax-by}$$