Respuesta :
Answer:
[tex]\tan \frac{13\pi}{12}=0.27[/tex]
Step-by-step explanation:
To find the exact value of [tex]\tan (195^{\circ})[/tex]
[tex]\tan (195^{\circ})[/tex] can be written in radian as [tex]\tan(\frac{13\pi}{12})[/tex]
Also,
[tex]\tan(\frac{13\pi}{12})=\tan(\frac{9\pi}{12}+\frac{4\pi}{12})[/tex]
Using formula for [tex]\tan A+ \tan B[/tex]
[tex]\tan A +\tan B =\frac{\tan A+\tan B}{1-\tan A\tan B}[/tex]
Here, [tex]A=\frac{3\pi}{4} \ and \ B=\frac{\pi}{4}[/tex]
Substitute,we have,
[tex]\tan \frac{13\pi}{12}=\dfrac{\tan \frac{3\pi}{4} +\tan \frac{\pi}{4}}{1-\tan \frac{3\pi}{4}\tan \frac{\pi}{4}}[/tex]
We know, [tex]\tan \frac{3\pi}{4}=-1 \ and\ \tan \frac{\pi}{4}=\sqrt{3}[/tex],
Putting, we get,
[tex]\tan \frac{13\pi}{12}=\dfrac{-1+\sqrt{3}}{1-(-1)\cdot \sqrt{3}}[/tex]
Solving, we get,
[tex]\tan \frac{13\pi}{12}=\dfrac{-1+\sqrt{3}}{1+\sqrt{3}}[/tex]
Rationalize by multiply and divide by [tex]1+\sqrt{3}[/tex] , we have,
[tex]\tan \frac{13\pi}{12}=\frac{-1+\sqrt{3}}{1+\sqrt{3}}\times\frac{1+\sqrt{3} }{1+\sqrt{3}}[/tex]
Simplify, we get,
[tex]\tan \frac{13\pi}{12}=2-\sqrt{3}[/tex]
Using , [tex]\sqrt{3}=1.73\\[/tex] we get,
[tex]\tan \frac{13\pi}{12}=0.27[/tex]
