Question

find the exact value of the following expression.

\\( \tan \left( \frac { \pi } { 6 } + \frac { 5 \pi } { 4 } \
ight) \\)

find the exact value of the expression.

\\( \tan \left( \frac { \pi } { 6 } + \frac { 5 \pi } { 4 } \
ight) = \square \\)

(simplify your answer. type an exact answer, using radicals as needed. use integers or fractions for any num

Explanation:

Step1: Use the tangent addition formula

The formula for \(\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\). Here \(A = \frac{\pi}{6}\) and \(B=\frac{5\pi}{4}\).
We know that \(\tan\frac{\pi}{6}=\frac{\sqrt{3}}{3}\) and \(\tan\frac{5\pi}{4}=\tan(\pi+\frac{\pi}{4})=\tan\frac{\pi}{4} = 1\) (since \(\tan(x + n\pi)=\tan x\) for \(n\in\mathbb{Z}\)).

Step2: Substitute the values into the formula

Substitute \(\tan A=\frac{\sqrt{3}}{3}\) and \(\tan B = 1\) into \(\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\).
We get \(\tan(\frac{\pi}{6}+\frac{5\pi}{4})=\frac{\frac{\sqrt{3}}{3}+ 1}{1-\frac{\sqrt{3}}{3}\times1}\).

Step3: Simplify the numerator and denominator

First, 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, \(\tan(\frac{\pi}{6}+\frac{5\pi}{4})=\frac{\frac{\sqrt{3}+3}{3}}{\frac{3 - \sqrt{3}}{3}}=\frac{\sqrt{3}+3}{3-\sqrt{3}}\).

Step4: Rationalize the denominator

Multiply the numerator and denominator by \(3+\sqrt{3}\).

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Answer:

\(2+\sqrt{3}\)