How to to a better approach for this :?

Question

If, $$x\cos A+y\sin A=k=x\cos B+y\sin B$$ Then find $(\cos A)(\cos B)$, $(\sin A)(\sin B)$ and $\cos A+\cos B$ and express them in terms of $x,y,k$ I found a solution but it included a really huge steps, Hope you can help me find an easy approach .

My "Bad Looking solution",

$x^2+y^2=(x^2+y^2)(sin^2A+cos^2A)=x^2sin^2A+y^2cos^2A-2xysinAcosA+x^2cos^2A+y^2sin^2A+2xysinAcosA$ Now it is easy to see that $x^2+y^2=(xsinA-ycosA)^2+k^2$

So, after many steps after finding the value of $sinA$ , $cosA$ from the simultaneous that we get, I got the ans, but it was really a very long way. :(

Answer 1

HINT:

So, $A,B$ are the roots of $$x\cos z+y\sin z=k$$

Now use Weierstrass substitution to form a Quadratic Eqaution in $\displaystyle\tan\frac z2$

So using Vieta's formula, we have $\displaystyle\tan\frac A2+\tan\frac B2, \tan\frac A2\tan\frac B2$ in terms of $\displaystyle k,x,y$

Can you take it home from here?

Answer 2

We have $\displaystyle x\cos A=k-y\sin A$

Squaring we get, $\displaystyle x^2\cos^2A=(k-y\sin A)^2\iff (x^2+y^2)\sin^2A-2ky\sin A+k^2-x^2=0$

We shall reach at the same Quadratic Equation if we start with $\displaystyle x\cos B+y\sin B=k\iff x\cos B=k-y\sin B$

So, $\displaystyle\sin A,\sin B$ are the roots of $\displaystyle(x^2+y^2)\sin^2C-2ky\sin C+k^2-x^2=0$

Now apply Vieta's formula

Can you follow the same for cosine

$\displaystyle x\cos B+y\sin B=k\iff y\sin B=k-x\cos B$ and sqaure