Given $\left(\;\sin(x-y)+\cos(x+2y)\sin y\;\right)^2=4\cos x\sin y\sin(x+y)$, prove $\tan x+\tan y=\cdots$

Question

Given $$\left(\;\sin(x-y)+\cos(x+2y)\sin y\;\right)^2=4\cos x\sin y\sin(x+y)$$ prove $$\tan x+\tan y=\frac{\tan y}{(\sqrt{2}\cos y-1)^2}$$

My attempt:

$$\left(\;\sin(x-y)+0.5\sin(x+3y)-0.5\sin x\;\right)^2=2\left(\;\sin(x+y)-\sin(x-y)\;\right)$$

I am not be able to prove last line. Help required.

Answer 1

Are known the trig identities \begin{align} &\sin(a\pm b) = \sin(a)\cos(b) \pm \cos(a)\sin(b),\tag1\\[4pt] &\cos(2a) = \cos^2(a) - \sin^2(a) = 2\cos^2(a) - 1 = 1-2\sin^2(a),\tag2\\[4pt] &\cos(a) - \cos(b) = 2\sin\dfrac{a+b}2\sin\dfrac{b-a}2.\tag3\\ \end{align} Using $(1) - (3),$ one can reduce the arguments of the LHS of the issue condition, \begin{align} &\sin(x-y)+\cos(x+2y)\sin (y) = \sin(x)\cos(y)+\cos(x+2y)\sin (y)-\cos(x)\sin(y) = \\[4pt] &\sin(x)\cos(y)-2\sin(x+y)\sin^2(y) = (1-2\sin^2(y))\sin(x+y)- \sin(y)\cos(x) =\\[4pt] &\cos(2y)\sin(x+y)- \sin(y)\cos(x).\\[4pt] \end{align}

Then \begin{align} &(\cos(2y)\sin(x+y)- \sin(y)\cos(x))^2 = 4\sin(y)\cos(x)\sin(x+y),\\ &\cos^2(2y)\sin^2(x+y)-2(\cos(2y)+2)\sin(y)\cos(x)\sin(x+y)+\sin^2(y)\cos^2(x) = 0.\tag4\\ \end{align}

Due to the area of admissible values of the goal equality, $\cos(x)\not=0$ and $\cos(y)\not=0.$ So it is correct to divide the equation $(4)$ by $\cos^2(x)\cos^2(y).$

Taking in account $(1)-(2),$ this leads to the square equation in the form of \begin{align} &\cos^2(2y)(\tan(x)+\tan(y))^2-2(\cos(2y)+2)\tan(y)(\tan(x)+\tan(y))^2+\tan^2(y) = 0,\\[4pt] &D = (\cos(2y)+2)^2 - \cos^2(2y) = 4(\cos(2y)+1) = 8\cos^2(y),\\[4pt] &\tan(x)+\tan(y) = \dfrac{\cos(2y)+2\pm\sqrt8\ \cos(y)}{\cos^2(2y)}\tan(y),\\[4pt] &\dfrac{\cos(2y)+2\pm2\sqrt2\ \cos(y)}{\cos^2(2y)} = \dfrac{2\cos^2(y)\pm2\sqrt2\ \cos(y) +1}{(2\cos^2(y)-1)^2} = \dfrac{\cos(2y)+2\pm\sqrt8\ \cos(y)}{(2\cos^2(y)-1)^2}\\[4pt] &= \dfrac{\left(\sqrt2\cos(y)\pm1\right)^2}{\left(\sqrt2\cos(y)+1\right)^2\left(\sqrt2\cos(y)-1\right)^2} = \dfrac1{\left(\sqrt2\cos(y)\pm1\right)^2},\\[4pt] &\tan(x)+\tan(y) = \dfrac{\tan(y)}{\left(\sqrt2\cos(y)\pm1\right)^2}.\tag5\\[4pt] \end{align} In accordance with the formula $(5),$ the initial problem setting looks incorrect, with a minimal discrepancy in the answers.

Maybe something is missing in the conditions?

Answer 2

You can approach this depending on your toolset. I prefer factoring expressions. Define $$ X := e^{ix}, \quad Y := e^{iy}, \quad A^- := \tan(x)+\tan(y)-\tan(y)/(1-\sqrt{2}\cos(y))^2, $$ $$ \quad A^+ := \tan(x)+\tan(y)-\tan(y)/(1+\sqrt{2}\cos(y))^2, $$ $$\quad C^- := 1 - \sqrt{2}Y(1+X) + XY^2, \quad C^+ := 1 + \sqrt{2}Y(1+X) + XY^2, $$ $$\quad D^- := 1 - \sqrt{2}Y(1-X) - XY^2, \quad D^+ := 1 + \sqrt{2}Y(1-X) - XY^2, $$ $$ B := (\sin(x-y)+\cos(x+2y)\sin(y))^2-4\cos(x)\sin(y)\sin(x+y),$$ where we used superscript ${^-}$ and ${^+}$ for $-\sqrt{2}$ and $+\sqrt{2},\;$ respectively. Using factoring we get that $$\;A^- = 2iC^-D^-/((1+X^2)(1-\sqrt{2}Y+Y^2)^2), \quad B = -C^-D^-C^+D^+ (1+Y^2)^2/(4XY^3)^2,\;$$ $$\;A^+ = 2iC^+D^+/((1+X^2)(1-\sqrt{2}Y+Y^2)^2).$$ If the numerator of $A^-$ is zero, then so is $B$. This proves $A^-=0$ implies $B=0.\;$ Similarly, if the numerator of $A^+$ is zero, then so is $B$. This proves $A^+=0$ implies $B=0.\;$ Finally, if $\;B=0\;$ then either $\;A^-=0\;$ or $\;A^+=0.$