A problem in trigonometry If  $\tan(x+y) =a+b$
and, $\tan(x-y)=a-b$
Then prove that:$a \tan(x) -b \tan(y)=a^{2}-b^{2}$
I  have used the formulae for $\tan(x+y)$ and $\tan(x-y)$ and then cross multiplied and then found $a$ and $b$ individually and then tried to put the values of $a$ and $b$ in "$a \tan(x) -b \tan(y)$"(l.h.s) but it did not lead to the r.h.s.
 A: In addition to David's: Let $u=\tan(x)$ and $v=\tan(y)$. You have
$$\frac{u+v}{a+b}=1-uv\quad\text{and}\quad\frac{u-v}{a-b}=1+uv.$$

What if you add two equations: $$\frac{u+v}{a+b}+\frac{u-v}{a-b}=1-uv+1+uv=2\\\implies(a-b)(u+v)+(a+b)(u-v)=2(a^2-b^2)\\ \cdots$$

A: hint....for simplicity write $x$ for $\tan x$ and $y$ for $\tan y$
then eliminate $xy$ from the pair of equations $$\frac{x+y}{1-xy}=a+b$$ and $$\frac{x-y}{1+xy}=a-b$$
A: We are given
$$
\tan(x+y)=a+b\tag1
$$
and
$$
\tan(x-y)=a-b\tag2
$$
Solving for $a$ gives
$$
\begin{align}
a
&=\frac12(\tan(x+y)+\tan(x-y))\\[3pt]
&=\frac12\left(\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}+\frac{\tan(x)-\tan(y)}{1+\tan(x)\tan(y)}\right)\\
&=\frac{\tan(x)+\tan(x)\tan^2(y)}{1-\tan^2(x)\tan^2(y)}\tag3
\end{align}
$$
and solving for $b$ gives
$$
\begin{align}
b
&=\frac12(\tan(x+y)-\tan(x-y))\\[3pt]
&=\frac12\left(\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}-\frac{\tan(x)-\tan(y)}{1+\tan(x)\tan(y)}\right)\\
&=\frac{\tan(y)+\tan^2(x)\tan(y)}{1-\tan^2(x)\tan^2(y)}\tag4
\end{align}
$$
Therefore,
$$
\begin{align}
a\tan(x)-b\tan(y)
&=\frac{\tan^2(x)-\tan^2(y)}{1-\tan^2(x)\tan^2(y)}\\
%&=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}\frac{\tan(x)-\tan(y)}{1+\tan(x)\tan(y)}\\[6pt]
%&=\tan(x+y)\tan(x-y)\\[9pt]
%&=a^2-b^2
\tag5
\end{align}
$$
Factoring the right hand side of $(5)$ leads quickly to the answer.
A: Hint:
$2a=a+b+(a-b)=?$
and
$2b=a+b-(a-b)=?$
