Solving for $\tan(A+B)$ with $(1 + \tan A)(1 + \tan B) = 2$
We are given the equation $(1 + \tan A)(1 + \tan B) = 2$ and need to find the value of $\tan(A+B)$.
Step-by-step Solution
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Expand the given equation:
$(1 + \tan A)(1 + \tan B) = 1 + \tan A + \tan B + \tan A \tan B$
So, $1 + \tan A + \tan B + \tan A \tan B = 2$
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Simplify the equation:
Subtract 1 from both sides:
$\tan A + \tan B + \tan A \tan B = 2 - 1$
$\tan A + \tan B + \tan A \tan B = 1$
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Rearrange the terms to match the tangent addition formula:
The formula for the tangent of the sum of two angles is:
$\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$
From the simplified equation in step 2, rearrange to isolate $\tan A + \tan B$:
$\tan A + \tan B = 1 - \tan A \tan B$
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Substitute into the tangent addition formula:
Divide both sides of $\tan A + \tan B = 1 - \tan A \tan B$ by $(1 - \tan A \tan B)$, assuming $1 - \tan A \tan B \neq 0$:
$\frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{1 - \tan A \tan B}{1 - \tan A \tan B}$
This simplifies to:
$\frac{\tan A + \tan B}{1 - \tan A \tan B} = 1$
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Identify the result:
The left side of the equation is the formula for $\tan(A+B)$.
Therefore, $\tan(A+B) = 1$
The value of $\tan(A+B)$ is 1.
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