4
$\begingroup$

How is it that $$\tan(A +B) = \frac{\tan A + \tan B}{1-\tan A\tan B}$$ for any value of $A$, $B$?

I have doubts about this since we arrive at this by dividing the numerator and denominator of $$\frac{\sin(A+B)}{\cos(A+B)}$$ by $\cos A \cdot \cos B$, which can only be done when $\cos A\cdot\cos B$ is not equal to zero.

  • 6
    tan(A+B) is **NOT** equal to (tanA+tanB)/(1-tanAtanB) when cos(A)cos(B)=0 since then tanA and/or tanB do not exist hence the RHS is not defined.2012-08-27
  • 0
    Strictly speaking, the tangent formula doesn't make sense when a cosine is zero, because when it is the tangent of that angle is undefined.2012-08-27
  • 5
    So it seems you just need to modify your statement to be "for any $A$ and $B$ which the expression is defined." This is usually implicit when talking about identities.2012-08-27
  • 0
    you need $A,B,A+B \neq \frac{\pi}{2}(2n-1)$ for $n \in \mathbb{Z}$ for the identity to hold.2012-08-27
  • 0
    I'd ask whether the identity makes sense if we define $\tan(\pm\pi/2) = \infty$, where there's just one $\infty$, at both ends of the line, rather than $+\infty\ne-\infty$. In other words, let $\tan$ take values in the projective line. I'll be surprised if that doesn't have an affirmative answer.2012-08-27

2 Answers 2