Hello Sallyvinka Originally Posted by
Sallyvinka Hi there,
I need help with proving two trigonometric expressions:
1) tanA - tanB = tanB tan(A-B)
tanA - cotB
As others have pointed out, there is a mistake in this question. It should read:$\displaystyle \frac{\tan A + \tan B}{\tan A - \cot B}=\tan B \tan(A-B)$
2)
tan (x - y) + tany
tan (x - y) - coty
You don't say what you want to prove here. But perhaps it's this:$\displaystyle \frac{\tan(x-y)+\tan y}{\tan(x-y) -\cot y}$$\displaystyle =\frac{\tan y(\tan(x-y)+\tan y)}{\tan y\tan(x-y) -1}$, multiplying 'top-and-bottom' by $\displaystyle \tan y$
$\displaystyle =-\tan y\cdot\frac{\tan(x-y)+\tan y}{1-\tan y\tan(x-y)}$, re-arranging
$\displaystyle =-\tan y \cdot\tan([x-y]+y)$, using $\displaystyle \tan(A+B) = \frac{\tan A + \tan B}{1-\tan A \tan B}$
$\displaystyle =-\tan x \tan y$
Grandad