Answer:
The answer is [tex]\bf\displaystyle\frac{\sqrt{3} }{3}[/tex].
Step-by-step explanation:
There are 2 approaches to solve question number 13:
- Since the given angles are special angles, we can easily find their values and do the calculation.
- Using the tan(α-β) formula
(1) Using the tangent value of [tex]\frac{4\pi}{3}[/tex] and [tex]\frac{7\pi}{6}[/tex]:
tangent value for special angles at Quadrant I [tex](0\leq \theta\leq \frac{\pi}{2} )[/tex]:
- [tex]tan(0)=0[/tex]
- [tex]tan(\frac{\pi}{6} )=\frac{\sqrt{3} }{3}[/tex]
- [tex]tan(\frac{\pi}{4} )=1[/tex]
- [tex]tan(\frac{\pi}{3} )=\sqrt{3}[/tex]
- [tex]tan(\frac{\pi}{2} )=\infty[/tex]
for tangent value at Quadrant II [tex](\frac{\pi}{2} \leq \theta\leq \pi)[/tex]:
- [tex]tan(\pi-\theta)=-tan(\theta)[/tex]
for tangent value at Quadrant III [tex](\pi\leq \theta\leq \frac{3\pi}{2} )[/tex]:
- [tex]tan(\pi+\theta)=tan(\theta)[/tex]
for tangent value at Quadrant IV [tex](\frac{3\pi}{2} \leq \theta\leq 2\pi)[/tex]:
- [tex]tan(2\pi-\theta)=-tan(\theta)[/tex]
First, we find the tangent value for [tex]\frac{4\pi}{3}[/tex] and [tex]\frac{7\pi}{6}[/tex]:
(i) [tex]\frac{4\pi}{3}[/tex] lies within Quadrant III, then
[tex]tan(\frac{4\pi}{3})=tan(\pi+\frac{\pi}{3} )[/tex]
[tex]=tan(\frac{\pi}{3} )[/tex]
[tex]=\sqrt{3}[/tex]
(i) [tex]\frac{7\pi}{6}[/tex] lies within Quadrant III, then
[tex]tan(\frac{7\pi}{6})=tan(\pi+\frac{\pi}{6} )[/tex]
[tex]=tan(\frac{\pi}{6} )[/tex]
[tex]=\frac{\sqrt{3}}{3}[/tex]
Therefore:
[tex]\displaystyle\frac{tan(\frac{4\pi}{3} )-tan(\frac{7\pi}{6} )}{1+tan(\frac{4\pi}{3} )tan(\frac{7\pi}{6} )} =\frac{\sqrt{3}-\frac{\sqrt{3} }{3} }{1+\sqrt{3}\left(\frac{\sqrt{3} }{3} \right)}[/tex]
[tex]\displaystyle=\frac{\sqrt{3}-\frac{\sqrt{3} }{3} }{1+1}[/tex]
[tex]\displaystyle=\frac{1}{2}\left(\sqrt{3}-\frac{\sqrt{3} }{3}\right)[/tex]
[tex]\displaystyle=\frac{1}{2}\left(\frac{3\sqrt{3}- \sqrt{3} }{3}\right)[/tex]
[tex]\displaystyle=\frac{1}{2}\left(\frac{2\sqrt{3} }{3}\right)[/tex]
[tex]\displaystyle=\frac{\sqrt{3} }{3}[/tex]
(2) Using the tan(α-β) formula:
[tex]\boxed{\displaystyle tan(\alpha-\beta)=\frac{tan(\alpha)-tan(\beta)}{1+tan(\alpha)tan(\beta)} }[/tex]
Let:
- [tex]tan(\alpha)=tan(\frac{4\pi}{3} )[/tex]
- [tex]tan(\beta)=tan(\frac{7\pi}{6} )[/tex]
Then:
[tex]\displaystyle \frac{tan(\alpha)-tan(\beta)}{1+tan(\alpha)tan(\beta)}=tan(\alpha-\beta)[/tex]
[tex]\displaystyle \frac{tan(\frac{4\pi}{3} )-tan(\frac{7\pi}{6})}{1+tan(\frac{4\pi}{3})tan(\frac{7\pi}{6})}=tan\left(\frac{4\pi}{3}-\frac{7\pi}{6}\right)[/tex]
[tex]\displaystyle=tan\left(\frac{8\pi-7\pi}{6} \right)[/tex]
[tex]\displaystyle=tan\left(\frac{\pi}{6} \right)[/tex]
[tex]\displaystyle=\frac{\sqrt{3} }{3}[/tex]
