Answer :
Answer:
To solve the equation \(\tan \beta + \tan 2\beta = \tan 3\beta\), we'll use trigonometric identities and properties of the tangent function.
### Step 1: Use the double-angle and triple-angle identities
We have the following trigonometric identities:
- Double-angle identity for tangent:
\[
\tan 2\beta = \frac{2 \tan \beta}{1 - \tan^2 \beta}
\]
- Triple-angle identity for tangent:
\[
\tan 3\beta = \frac{3 \tan \beta - \tan^3 \beta}{1 - 3 \tan^2 \beta}
\]
### Step 2: Substitute the identities into the original equation
Given:
\[
\tan \beta + \tan 2\beta = \tan 3\beta
\]
Substitute \(\tan 2\beta\) and \(\tan 3\beta\):
\[
\tan \beta + \frac{2 \tan \beta}{1 - \tan^2 \beta} = \frac{3 \tan \beta - \tan^3 \beta}{1 - 3 \tan^2 \beta}
\]
### Step 3: Simplify the equation
Let's simplify the left side first:
\[
\tan \beta + \frac{2 \tan \beta}{1 - \tan^2 \beta} = \frac{\tan \beta (1 - \tan^2 \beta) + 2 \tan \beta}{1 - \tan^2 \beta} = \frac{\tan \beta (1 - \tan^2 \beta + 2)}{1 - \tan^2 \beta} = \frac{\tan \beta (3 - \tan^2 \beta)}{1 - \tan^2 \beta}
\]
Now, the equation becomes:
\[
\frac{\tan \beta (3 - \tan^2 \beta)}{1 - \tan^2 \beta} = \frac{3 \tan \beta - \tan^3 \beta}{1 - 3 \tan^2 \beta}
\]
### Step 4: Equate the numerators and the denominators
To solve the equation, we set the numerators and denominators equal to each other, as long as neither denominator is zero:
\[
\tan \beta (3 - \tan^2 \beta) = 3 \tan \beta - \tan^3 \beta
\]
Since \(\tan \beta \neq 0\), we can divide both sides by \(\tan \beta\):
\[
3 - \tan^2 \beta = 3 - \tan^2 \beta
\]
This equation is always true, indicating that the given equation holds for all values of \(\beta\) for which the tangent function and the angle identities are defined. Therefore, there is no specific solution for \(\beta\); rather, the original equation is an identity that holds for all values of \(\beta\) where the tangent function is defined and where the denominators are non-zero.
Answer:I don’t really like maths so I don’t know the answer
Step-by-step explanation:so yeah