Question

find the exact value of the trigonometric expression without the use of a calculator.
find the exact value of the expression.
$\tan\left(\frac{7\pi}{6}-\frac{7\pi}{4}\
ight)=\square$
(simplify your answer. type an exact answer, using radicals as neede

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 \(3 + \sqrt{3}\).

$$
LATEXBLOCK0
$$

Answer:

\(2+\sqrt{3}\)