Use a sum or difference formula to find the exact value of the following (tanfrac (3pi )/(4)+tan(pi )/(12))(1-tan(3pi )/(4)tan(pi )/(12)) square

To find the exact value of the given expression, we can use the tangent sum formula:<br /><br />$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$<br /><br />In this case, we have $A = \frac{3\pi}{4}$ and $B = \frac{\pi}{12}$.<br /><br />First, let's find the values of $\tan\frac{3\pi}{4}$ and $\tan\frac{\pi}{12}$.<br /><br />$\tan\frac{3\pi}{4} = -1$<br /><br />To find $\tan\frac{\pi}{12}$, we can use the tangent subtraction formula:<br /><br />$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$<br /><br />Let $A = \frac{\pi}{4}$ and $B = \frac{\pi}{6}$, then $\frac{\pi}{12} = \frac{\pi}{4} - \frac{\pi}{6}$.<br /><br />$\tan\frac{\pi}{12} = \tan(\frac{\pi}{4} - \frac{\pi}{6}) = \frac{\tan\frac{\pi}{4} - \tan\frac{\pi}{6}}{1 + \tan\frac{\pi}{4} \tan\frac{\pi}{6}}$<br /><br />Since $\tan\frac{\pi}{4} = 1$ and $\tan\frac{\pi}{6} = \frac{\sqrt{3}}{3}$, we have:<br /><br />$\tan\frac{\pi}{12} = \frac{1 - \frac{\sqrt{3}}{3}}{1 + 1 \cdot \frac{\sqrt{3}}{3}} = \frac{3 - \sqrt{3}}{3 + \sqrt{3}}$<br /><br />Now, substitute the values of $\tan\frac{3\pi}{4}$ and $\tan\frac{\pi}{12}$ into the original expression:<br /><br />$\frac{\tan\frac{3\pi}{4} + \tan\frac{\pi}{12}}{1 - \tan\frac{3\pi}{4} \tan\frac{\pi}{12}} = \frac{-1 + \frac{3 - \sqrt{3}}{3 + \sqrt{3}}}{1 - (-1) \cdot \frac{3 - \sqrt{3}}{3 + \sqrt{3}}}$<br /><br />Simplify the numerator and denominator separately:<br /><br />Numerator: $-1 + \frac{3 - \sqrt{3}}{3 + \sqrt{3}} = \frac{-(3 + \sqrt{3}) + (3 - \sqrt{3})}{3 + \sqrt{3}} = \frac{-3 - \sqrt{3} + 3 - \sqrt{3}}{3 + \sqrt{3}} = \frac{-2\sqrt{3}}{3 + \sqrt{3}}$<br /><br />Denominator: $1 - (-1) \cdot \frac{3 - \sqrt{3}}{3 + \sqrt{3}} = 1 + \frac{3 - \sqrt{3}}{3 + \sqrt{3}} = \frac{3 + \sqrt{3} + 3 - \sqrt{3}}{3 + \sqrt{3}} = \frac{6}{3 + \sqrt{3}}$<br /><br />Now, divide the numerator by the denominator:<br /><br />$\frac{\frac{-2\sqrt{3}}{3 + \sqrt{3}}}{\frac{6}{3 + \sqrt{3}}} = \frac{-2\sqrt{3}}{6} = -\frac{\sqrt{3}}{3}$<br /><br />Therefore, the exact value of the given expression is $-\frac{\sqrt{3}}{3}$.