Question

find the exact value of the following expression.
tan(11π/6 + 3π/4)
find the exact value of the expression.
tan(11π/6 + 3π/4) = □
(simplify your answer. type an exact answer, using radicals as needed. use integers or fractions for expression. rationalize all denominators.)

Explanation:

Step1: Use the tangent addition formula

The tangent addition formula is $\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$. Let $A = \frac{11\pi}{6}$ and $B=\frac{3\pi}{4}$.
First, find $\tan\frac{11\pi}{6}$ and $\tan\frac{3\pi}{4}$.
We know that $\tan\frac{11\pi}{6}=\tan(2\pi-\frac{\pi}{6})=-\tan\frac{\pi}{6}=-\frac{\sqrt{3}}{3}$, and $\tan\frac{3\pi}{4}=- 1$.

Step2: Substitute into the formula

Substitute $\tan A=-\frac{\sqrt{3}}{3}$ and $\tan B = - 1$ into the formula $\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$.
We get $\tan(\frac{11\pi}{6}+\frac{3\pi}{4})=\frac{-\frac{\sqrt{3}}{3}+(-1)}{1-(-\frac{\sqrt{3}}{3})\times(-1)}$.
Simplify the numerator: $-\frac{\sqrt{3}}{3}-1=-\frac{\sqrt{3}+3}{3}$.
Simplify the denominator: $1-\frac{\sqrt{3}}{3}=\frac{3 - \sqrt{3}}{3}$.
So, $\tan(\frac{11\pi}{6}+\frac{3\pi}{4})=\frac{-\frac{\sqrt{3}+3}{3}}{\frac{3-\sqrt{3}}{3}}=\frac{-(\sqrt{3}+3)}{3 - \sqrt{3}}$.

Step3: Rationalize the denominator

Multiply the numerator and denominator by the conjugate of the denominator $3+\sqrt{3}$.

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Answer:

$-2-\sqrt{3}$