For 2 orthogonal vectors, is tan(a)tan(b)=-1? 
This is a part of a physics subproblem I was solving where in the solution they casually mentioned:-
$\begin{align}\\
tan(A) \times tan(B) = -1
\end{align}\\$
Without proving it, and proceeded to solve the complete question. I tried using ASS similarity methods but it yielded a totally different result.
I am willing to post my physics question here in case someone thinks this is an XY scenario or I got something wrong! :]

Use the definition $\tan(x) \triangleq \frac{\sin(x)}{\cos(x)}$ and trigonometric product identities (Google them).
$\begin{align}
\tan(A) \tan(B) &= \frac{\sin(A)}{\cos(A)} \frac{\sin(B)}{\cos(B)}\\
&= \frac{\sin(A) \sin(B)}{\cos(A)\cos(B)}\\
&= \frac{\frac{1}{2} \left(\cos(A - B) - \cos(A + B) \right) }{\frac{1}{2} \left(\cos(A + B) + \cos(A - B) \right)}\\
&= \frac{\frac{1}{2} \left(\cos(A - B) - \cos(90) \right) }{\frac{1}{2} \left(\cos(90) + \cos(A - B) \right)}\\
&= \frac{\frac{1}{2} \left(\cos(A - B) - 0 \right) }{\frac{1}{2} \left(0 + \cos(A - B) \right)}\\
&= 1
\end{align}$
If the angles are taken to be signed angles, then the above proof stills works using the additional fact that $\tan(-B) = - \tan(B)$. This means that angle A is positive/anticlockwise and angle B is negative/clockwise.
