# To prove a trigonometric identity with tan() and cot()

brotherbobby
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.

• Delta2

Homework Helper
Gold Member
2022 Award
Why didn't you use equation 2) from your relevant equations?

brotherbobby

Let me finish my argument above nonetheless.

Continuing from above (Post#1)

\begin{align*}
& = 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}\\
& = 1+ \frac{\sec^2 A \frac{\sin A}{\cos B \cos C}+\sec^2 B \frac{\sin B}{\cos C \cos A}+\sec^2 C \frac{\sin C}{\cos A \cos B}}{\tan A \tan B \tan C}\; \small{\text{[using}\;(\sin B \cos C+\cos B\sin C = \sin (B+C)=\sin (\pi -A) = \sin A\; \text{etc.})]}\\
&= 1+ \frac{\frac{\tan A+\tan B+\tan C}{\cos A \cos B \cos C}}{\tan A \tan B \tan C}\; \small{[\text{taking cos A cos B cos C as common LCM}]}\\
&=\boxed{1+\sec A \sec B \sec C} \; \small{[\text{Using}\; \tan A +\tan B +\tan C = \tan A \tan B \tan C]}\\
\end{align*}