- #1

brotherbobby

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- Homework Statement
- If angles ##A+B+C=\pi##, prove that ##(\tan A+\tan B+\tan C)(\cot A+\cot B+\cot C) = \boxed{1+\sec A \sec B \sec C}##

- Relevant Equations
- 1. The tangents's and cosine's are reciprocals, e.g. ##\tan x \cot x = 1##

2. If ##A+B+C = \pi\Rightarrow \tan A+\tan B+\tan C = \tan A \tan B \tan C##

**Attempt : I could not progress far, but the following is what I could do.**

$$\begin{align*}

\mathbf{\text{LHS}} & = (\tan A+\tan B+\tan C)(\cot A+\cot B+\cot C) \\

& = 3+\tan A \cot B+\tan B \cot A+\tan A \cot C+\tan C \cot A+\tan B \cot C+\tan C \cot B\\

& = 3+\frac{\tan^2A+\tan^2B}{\tan A \tan B}+\frac{\tan^2C+\tan^2A}{\tan C \tan A}+\frac{\tan^2B+\tan^2C}{\tan B \tan C}\\

& = 3+\frac{\tan^2A (\tan B+\tan C) + \tan^2B (\tan C+\tan A)+\tan^2C (\tan A+\tan B)}{\tan A \tan B \tan C}\\

& = 3+\frac{(\sec^2A-1) (\tan B+\tan C)+(\sec^2B-1) (\tan C+\tan A)+(\sec^2C-1) (\tan A+\tan B)}{\tan A \tan B \tan C}\\

& = 3+\frac{(\sec^2A) (\tan B+\tan C)+(\sec^2B) (\tan C+\tan A)+(\sec^2C) (\tan A+\tan B)-2\overbrace{(\tan A+\tan B+\tan C)}^{\tan A \tan B \tan C}}{\tan A \tan B \tan C}\\

& \\

& = 1 + \frac{(\sec^2A) (\tan B+\tan C)+(\sec^2B) (\tan C+\tan A)+(\sec^2C) (\tan A+\tan B)}{\tan A \tan B \tan C}\\

\end{align*}$$

The numerator has to end up being ##\sec A \tan A \sec B \tan B \sec C \tan C## in order to get the answer. I have got the "+1" in the first term.

**I do not know how to proceed from here. Some help in the form of hints would be welcome.**