Questions: Which expression is equivalent to tan(x-π)? (tan(x)-tan(π))/(1-(tan(x))(tan(π))) (tan(x)-tan(π))/(1+(tan(x))(tan(π))) (tan(x)+tan(π))/(1-(tan(x))(tan(π))) (tan(x)+tan(π))/(1+(tan(x))(tan(π)))

Transcript text: Which expression is equivalent to $\tan (x-\pi)$ ? $\frac{\tan (x)-\tan (\pi)}{1-(\tan (x))(\tan (\pi))}$ $\frac{\tan (x)-\tan (\pi)}{1+(\tan (x))(\tan (\pi))}$ $\frac{\tan (x)+\tan (\pi)}{1-(\tan (x))(\tan (\pi))}$ $\frac{\tan (x)+\tan (\pi)}{1+(\tan (x))(\tan (\pi))}$

Solution

Solution Steps

To find an expression equivalent to $\tan(x - \pi)$, we can use the tangent subtraction formula: $\tan(a - b) = \frac{\tan(a) - \tan(b)}{1 + \tan(a)\tan(b)}$. Here, $a = x$ and $b = \pi$. We know that $\tan(\pi) = 0$, so we substitute these values into the formula to simplify the expression.

Step 1: Use the Tangent Subtraction Formula

To find an expression equivalent to $\tan(x - \pi)$, we apply the tangent subtraction formula: \[ \tan(a - b) = \frac{\tan(a) - \tan(b)}{1 + \tan(a)\tan(b)} \] where $a = x$ and $b = \pi$.

Step 2: Substitute Values

Substituting $a$ and $b$ into the formula, we have: \[ \tan(x - \pi) = \frac{\tan(x) - \tan(\pi)}{1 + \tan(x)\tan(\pi)} \] Since $\tan(\pi) = 0$, the expression simplifies to: \[ \tan(x - \pi) = \frac{\tan(x) - 0}{1 + \tan(x) \cdot 0} = \frac{\tan(x)}{1} = \tan(x) \]

Step 3: Conclusion

Thus, we find that: \[ \tan(x - \pi) = \tan(x) \]