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

• brotherbobby
In summary, the conversation is about solving an equation and reaching an answer of 1+sec A sec B sec C. The individual tried different approaches and eventually reached the solution. They also mention using relevant equations and receiving help in the form of hints.
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
Why didn't you use equation 2) from your relevant equations?

I have got the answer. Sorry to waste your time *blush*.

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*}

## 1. What is a trigonometric identity?

A trigonometric identity is an equation that is true for all values of the variables involved. In other words, it is a statement that is always true, regardless of the specific values used for the trigonometric functions.

## 2. How do you prove a trigonometric identity with tan() and cot()?

To prove a trigonometric identity with tan() and cot(), you will need to use the basic trigonometric identities and algebraic manipulation. You will also need to use the definitions of tan() and cot() in terms of sine and cosine.

## 3. What are the basic trigonometric identities?

The basic trigonometric identities include the Pythagorean identities, reciprocal identities, quotient identities, and even/odd identities. These identities relate the trigonometric functions to each other and are essential for proving more complex identities.

## 4. Can you give an example of proving a trigonometric identity with tan() and cot()?

Sure, for example, let's say we want to prove the identity tan(x) + cot(x) = sec(x)csc(x). We can start by writing tan(x) and cot(x) in terms of sine and cosine, and then use the reciprocal and quotient identities to simplify the expression until it is equivalent to sec(x)csc(x).

## 5. Why is it important to know how to prove trigonometric identities?

Proving trigonometric identities is important because it allows us to verify the relationships between different trigonometric functions and can be used to solve more complex trigonometric equations. It also helps to deepen our understanding of the fundamental concepts of trigonometry.

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