Proving $a\, \tan\,x - b\, \tan\,y = a^2 - b^2$ with $\tan(x+y)$ & $\tan(x-y)$

[kaliprasad]

If $\tan(x+y) = a + b$ and $\tan(x-y) = a - b$ show that $a\, \tan\,x - b\, \tan\,y = a^2 - b^2$

 

[lfdahl]

My solution:

Spoiler

\[\left\{\begin{matrix} x+y = \arctan (a+b)\\ x-y = \arctan (a-b) \end{matrix}\right.\Rightarrow \left\{\begin{matrix} 2x = \arctan (a+b) + \arctan (a-b) = \arctan \left ( \frac{2a}{1-(a^2-b^2)} \right ) \\ 2y = \arctan (a+b) - \arctan (a-b) = \arctan \left ( \frac{2b}{1+(a^2-b^2)} \right ) \end{matrix}\right.\]

\[ \Rightarrow \left\{\begin{matrix} \tan(2x) = \frac{2\tan x}{1-\tan ^2x}= \frac{2a}{1-(a^2-b^2)}\\ \tan(2y) = \frac{2\tan y}{1-\tan ^2y}= \frac{2b}{1+(a^2-b^2)} \end{matrix}\right. \Rightarrow \left\{\begin{matrix} a \tan x = \frac{a^2}{1-(a^2-b^2)}(1-\tan ^2x)\\ b \tan y = \frac{b^2}{1+(a^2-b^2)}(1-\tan ^2y) \end{matrix}\right.\]

\[ \Rightarrow \left\{\begin{matrix} (a\tan x)^2+(1-(a^2-b^2))(a\tan x) -a^2 = 0\\ (b\tan y)^2+(1+(a^2-b^2))(b\tan y) -b^2 = 0 \end{matrix}\right. \]

\[ \Rightarrow \left\{\begin{matrix} a\tan x = \frac{1}{2}\left ( -(1-(a^2-b^2)) \pm \sqrt{(1-(a^2-b^2))^2+4a^2} \right )\\ b\tan x = \frac{1}{2}\left ( -(1+(a^2-b^2)) \pm \sqrt{(1+(a^2-b^2))^2+4b^2} \right ) \end{matrix}\right.\] \[ \Rightarrow \left\{\begin{matrix} a \tan x = \frac{1}{2}\left ( -(1-(a^2-b^2)) \pm \sqrt{1+(a^2-b^2)^2+ 2(a^2+b^2)}\right )\\ b \tan y = \frac{1}{2}\left ( -(1+(a^2-b^2)) \pm \sqrt{1+(a^2-b^2)^2+ 2(a^2+b^2)}\right ) \end{matrix}\right. \] \[ \Rightarrow a \tan x - b \tan y = \frac{1}{2}(-1+(a^2-b^2)+1+(a^2-b^2)) = a^2-b^2.\]

 

[kaliprasad]

My Solution:

Spoiler

$\tan(x+y) = \frac{\tan\,x+\tan\,y}{1-\tan\,x\tan\,y}$ or $(a+b)(1-\tan\,x\tan\,y) = \tan\,x+\tan\,y \cdots(1)$
Similarly $(a-b)(1+\tan\,x\tan\,y) = \tan\,x-\tan\,y\cdots(2)$
multiplying (1) by (a-b) and (2) by (a+b) and adding we get $(a^2-b^2) = 2a\tan\,x-2b\tan \,y$
divide both sides by 2 to get the result.