Hello Punch Originally Posted by
Punch It is all so confusing... I have tried to combine the 2 posts but it makes no sense to me.
If $\displaystyle A+B+C=90^\circ$, show that $\displaystyle tanAtanB+tanBtanC+tanCtanA=1$
How about telling me the idea of solving this question? Is it about changing everything in terms of A?
I'm not sure what you mean by this. You were asked to show that if a certain statement was true (namely, $\displaystyle A+B+C=90^o$) then another statement follows (namely, $\displaystyle \tan A\tan B+\tan B\tan C+\tan C\tan A=1$).
I have done this by starting with the equation $\displaystyle A+B+C=90^o$ and ending with the equation $\displaystyle \tan A\tan B+\tan B\tan C+\tan C\tan A=1$.
Putting together my two posts, we get:$\displaystyle A+B+C=90^o$
$\displaystyle \Rightarrow A+B = 90^o-C$
$\displaystyle \Rightarrow \tan(A+B) = \tan(90^o-C)$
$\displaystyle \Rightarrow \frac{\tan A + \tan B}{1-\tan A\tan B}= \cot C$$\displaystyle =\frac{1}{\tan C}$
$\displaystyle \Rightarrow \tan C(\tan A + \tan B) = 1 - \tan A \tan B$
$\displaystyle \Rightarrow \tan A\tan B + \tan B\tan C +\tan C \tan A = 1$
So we have now done everything we were asked to do. Are there any steps in the working that you don't understand?
Grandad