find the exact value of the following expression.\n\n\\( \\tan \\left( \\frac { \\pi } { 6 } + \\frac { 5…

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}). [ \begin{align*} \frac{\sqrt{3}+3}{3-\sqrt{3}}\times\frac{3+\sqrt{3}}{3+\sqrt{3}}&=\frac{(\sqrt{3}+3)(\sqrt{3}+3)}{(3-\sqrt{3})(3 + \sqrt{3})}\ &=\frac{3 + 6\sqrt{3}+9}{9-3}\ &=\frac{12 + 6\sqrt{3}}{6}\ &=2+\sqrt{3} \end{align*} ]

Answer:

(2+\sqrt{3})