Respuesta :

With θ in quadrant III, we expect to have both cosθ and sinθ negative. Then from the Pythagorean identity, we get

[tex]\cos^2\theta+\sin^2\theta=1\implies\sin\theta=-\sqrt{1-\cos^2\theta}=-\dfrac{\sqrt{35}}6[/tex]

Then from the definition of tangent, that is

tanθ = sinθ / cosθ

we get

[tex]\tan\theta=\dfrac{-\frac{\sqrt{35}}6}{-\frac16}=\boxed{\sqrt{35}}[/tex]

The value of [tex]\tan \theta[/tex] is required.

The required value in quadrant three is [tex]\tan\theta=\sqrt{35}[/tex]

[tex]\cos \theta=-\dfrac{1}{6}[/tex]

We have the identity

[tex]\tan \theta=\dfrac{\sqrt{1-\cos^2\theta}}{\cos \theta}\\\Rightarrow \tan\theta=\dfrac{\sqrt{1-\left(-\dfrac{1}{6}\right)^2}}{-\dfrac{1}{6}}\\\Rightarrow \tan \theta=\dfrac{\sqrt{\dfrac{35}{36}}}{-\dfrac{1}{6}}\\\Rightarrow \tan\theta=\pm\sqrt{35}\times -1\\\Rightarrow \tan\theta=\pm\sqrt{35}[/tex]

The required value in quadrant three is [tex]\tan\theta=\sqrt{35}[/tex]

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