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info@snapandsolve.comTo find the exact value of \( \tan \left( \frac{5\pi}{6} + \frac{3\pi}{4} \right) \), we use the tangent addition formula:
\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]
First, calculate \( \tan \left( \frac{5\pi}{6} \right) \) and \( \tan \left( \frac{3\pi}{4} \right) \):
Now apply the formula:
\[ \tan \left( \frac{5\pi}{6} + \frac{3\pi}{4} \right) = \frac{-\frac{1}{\sqrt{3}} + (-1)}{1 - \left(-\frac{1}{\sqrt{3}}\right)(-1)} \]
\[ = \frac{-\frac{1}{\sqrt{3}} - 1}{1 - \frac{1}{\sqrt{3}}} \]
\[ = \frac{-\frac{1 + \sqrt{3}}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} \]
\[ = \frac{-\frac{1 + \sqrt{3}}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} \]
\[ = -\frac{1 + \sqrt{3}}{\sqrt{3} - 1} \]
Rationalize the denominator:
\[ = -\frac{(1 + \sqrt{3})(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} \]
\[ = -\frac{(1 + \sqrt{3})(\sqrt{3} + 1)}{3 - 1} \]
\[ = -\frac{(1 + \sqrt{3})(\sqrt{3} + 1)}{2} \]
\[ = -\frac{(\sqrt{3} + 1 + 3 + \sqrt{3})}{2} \]
\[ = -\frac{4 + 2\sqrt{3}}{2} \]
\[ = -2 - \sqrt{3} \]
Thus, the exact value is \(-2 - \sqrt{3}\).
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