4

I was studying conditional identities for triangle in trigonometry where I had to prove that

$\tan{A} + \tan{B} + \tan{C} = \tan{A}\tan{B}\tan{C}$

So I started with $\tan({A+B+C)}=\frac{S1-S3}{1-S2}$ where $S1=\sum_{cyc}\tan A$, $S2=\sum_{cyc}\tan A.\tan B$ and $S3=\prod_{cyc}\tan A$

Now since $A+B+C=\pi$ and since $\tan\pi=0$, therefore $\frac{S1-S3}{1-S2}=0$ and hence $S1=S3$ but I got stuck when I realized that another condition, $1-S2\neq0$ should also be true. Now I can't prove this.

I'm not having any idea how to start to prove $S2\neq1$ for a triangle. please give me a hint.

Calvin Khor
  • 36,192
  • 6
  • 47
  • 102
Lalit Tolani
  • 3,454

2 Answers2

4

$$\cos (A+B+C) = \cos A \cos B \cos C \left(1-\sum_{cyc}\tan A \tan B\right)$$

since $A+B+C=\pi$ in a triangle,

$$\cos \pi=\cos A \cos B \cos C \left(1-\sum_{cyc}\tan A \tan B\right)$$

$$\implies \cos A \cos B \cos C \left(1-\sum_{cyc}\tan A \tan B \right)=-1$$

if $S_{2}=1$,the LHS of the above equation will be zero , so , $S_{2}\neq1$

IITM
  • 450
2

$~~~~\tan A\tan B+\tan B\tan C+\tan A\tan C\\=\tan A \cdot [\tan (B+C) \cdot (1-\tan B ~tan C)]+\tan B\tan C\\ =-(\tan A)^2 \cdot (1-\tan B ~tan C)+\tan B\tan C\\ =(\tan^2 A+1)(\tan B \tan C)-\tan^2 A$

Assume there exist solution for $~\tan A\tan B+\tan B\tan C+\tan A\tan C=1$

$⇒ (\tan^2 A+1)(\tan B \tan C)=\tan^2 A+1$
$⇒\tan B \tan C=1$
$~~~~~\tan A=-\tan (B+C)=\frac{\tan B+\tan C}{1-\tan B \tan C}~~$ which is not defined $($as denominator $= 0)$.

Hence $~\tan A\tan B+\tan B\tan C+\tan A\tan C \ne 1$

Asher2211
  • 3,416
  • tanA is not defined therefore A can be 90 – Lalit Tolani Jun 01 '21 at 03:02
  • @LalitTolani But you can't substitute $\tan A$ in $~\tan A\tan B+\tan B\tan C+\tan A\tan C $ – Asher2211 Jun 01 '21 at 03:05
  • why we can't substitute – Lalit Tolani Jun 01 '21 at 03:08
  • $\tan A$ is not defined at $A=\frac{\pi}{2}$. You can't substitute something that is not defined. If you take $ A \rightarrow \frac{\pi}{2}$ then $\tan A\tan B+\tan B\tan C+\tan A\tan C \ne 1$ – Asher2211 Jun 01 '21 at 03:13
  • that means you want to say that S2=1 gives an undefined value for tanA . I understood till here then what – Lalit Tolani Jun 01 '21 at 03:20
  • When $A \rightarrow \frac{\pi}{2}$ then $\tan A\tan B+\tan B\tan C+\tan A\tan C \rightarrow +\infty$ or $-\infty$(Depending on whether $\tan B$ is positive or negative).Note that $\tan B$ and $\tan C$ can't be $0$ at the same time and $\tan B$ and $\tan C$ have the same sign because $\tan B=\tan (\frac{\pi}{2}-C)=\frac{1}{\tan C}$. – Asher2211 Jun 01 '21 at 03:28
  • So now I understood S2=1 when solved gives an undefined value of S2 and since undefined can never be equal to something defined therefore S2 should be not equal to 1 – Lalit Tolani Jun 01 '21 at 03:31