find the exact value of the trigonometric expression without the use of a calculator.\nfind the exact value…

Explanation:

Step1: Use the tangent subtraction formula

The formula for (\tan(A - B)=\frac{\tan A-\tan B}{1 + \tan A\tan B}). Here (A=\frac{7\pi}{6}) and (B = \frac{7\pi}{4}). First, find (\tan\frac{7\pi}{6}) and (\tan\frac{7\pi}{4}). For (\tan\frac{7\pi}{6}), (\frac{7\pi}{6}=\pi+\frac{\pi}{6}). Using the identity (\tan(\pi+\alpha)=\tan\alpha), so (\tan\frac{7\pi}{6}=\tan(\pi+\frac{\pi}{6})=\tan\frac{\pi}{6}=\frac{\sqrt{3}}{3}). For (\tan\frac{7\pi}{4}), (\frac{7\pi}{4}=2\pi-\frac{\pi}{4}). Using the identity (\tan(2\pi-\alpha)=-\tan\alpha), so (\tan\frac{7\pi}{4}=\tan(2\pi - \frac{\pi}{4})=-\tan\frac{\pi}{4}=- 1).

Step2: Substitute 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{7\pi}{6}-\frac{7\pi}{4})=\frac{\frac{\sqrt{3}}{3}-(-1)}{1+\frac{\sqrt{3}}{3}\times(-1)}=\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 (3 + \sqrt{3}). [ \begin{align*} \frac{\sqrt{3}+3}{3-\sqrt{3}}\times\frac{3+\sqrt{3}}{3+\sqrt{3}}&=\frac{(\sqrt{3}+3)^{2}}{(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})