find the exact value of the following expression.\n\n(\tan left(\frac{pi}{6}+\frac{5 pi}{4}\right))\n\nfind…

Explanation:

Step1: Use the tangent addition formula

The tangent addition formula is $\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$. Here $A = \frac{\pi}{6}$ and $B=\frac{5\pi}{4}$. We know that $\tan\frac{\pi}{6}=\frac{\sqrt{3}}{3}$ and $\tan\frac{5\pi}{4}=\tan(\pi+\frac{\pi}{4})=\tan\frac{\pi}{4} = 1$. So, $\tan(\frac{\pi}{6}+\frac{5\pi}{4})=\frac{\tan\frac{\pi}{6}+\tan\frac{5\pi}{4}}{1 - \tan\frac{\pi}{6}\tan\frac{5\pi}{4}}$.

Step2: Substitute the values

Substitute $\tan\frac{\pi}{6}=\frac{\sqrt{3}}{3}$ and $\tan\frac{5\pi}{4} = 1$ into the formula: [ \begin{align*} \frac{\frac{\sqrt{3}}{3}+1}{1-\frac{\sqrt{3}}{3}\times1}&=\frac{\frac{\sqrt{3}+ 3}{3}}{\frac{3-\sqrt{3}}{3}}\ &=\frac{\sqrt{3}+3}{3-\sqrt{3}} \end{align*} ]

Step3: Rationalize the denominator

Multiply the numerator and denominator by the conjugate of the denominator $3 + \sqrt{3}$: [ \begin{align*} \frac{(\sqrt{3}+3)(3+\sqrt{3})}{(3-\sqrt{3})(3+\sqrt{3})}&=\frac{3\sqrt{3}+3 + 9+3\sqrt{3}}{9-3}\ &=\frac{6\sqrt{3}+12}{6}\ &=\sqrt{3}+ 2 \end{align*} ]

Answer:

$\sqrt{3}+2$