System of Trigonomtric Equations

Question

Please Help to solve the following problem :- If, $$\sin(x+y)=a$$ $$\cos(x-y)=b$$ $$a,b\in \Bbb R^+$$ Find $\tan(2x)$ in terms of a and b.

Answer 1

$$ x+y = \arcsin a\\ x-y = \arccos b $$

then

$$ 2x = \arcsin a + \arccos b \to \tan(2x) = \frac{a\sqrt{1-a^2}+b \sqrt{1-b^2}}{b^2-a^2} $$

NOTE

$$ \tan(u+v) = \frac{\sin (u) \cos (v)}{\cos (u) \cos (v)-\sin (u) \sin (v)}+\frac{\cos (u) \sin (v)}{\cos (u) \cos (v)-\sin (u) \sin (v)} $$

Answer 2

$\cos2x=\cos(x+y+x-y)=\cos(x+y)cos(x-y)-\sin(x+y)\sin(x-y)= a\sqrt{1-b^2}- b\sqrt{1-a^2}$.

Similarly $\sin2x=\sin(x+y+x-y)=\sin(x+y)\cos(x-y)+\cos(x+y)\sin(x-y)=\sqrt{1-a^2}\sqrt{1-b^2}+ab$.

Thus...?

Note: this works for quadrant $1$. In $2,3$ and $4$ you need to change some signs.