Given the identity:
![sin^2\theta+cos^2\theta=1](https://img.qammunity.org/2023/formulas/mathematics/college/4l3jle1tms393h00pwlwcixxjikhkyvr92.png)
Let's find the value of tanθ, where:
![cos\theta=-0.27\text{ and }\pi<\theta<(3\pi)/(2)](https://img.qammunity.org/2023/formulas/mathematics/college/u2krbf99laahme1cut390ocyie6uoprgdb.png)
Since it is in the given interval, θ is in the third quadrant.
Now, substitute -0.27 for θ in the identity:
![\begin{gathered} sin^2\theta+(-0.27)^2=1 \\ \\ sin^2\theta+0.0729=1 \\ \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/g75jsnx2sssa2vsv01dfgpaq1oli32axad.png)
Subtract 0.0729 from both sides:
![\begin{gathered} sin^2\theta+0.0729-0.0729=1-0.0729 \\ \\ sin^2\theta=0.9271 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/imt6i1r0q4xs2t9h99g84dp6839pdxhqk7.png)
Take the square root of both sides:
![\begin{gathered} sin\theta=√(0.9271) \\ \\ sin\theta=0.9629 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/yka81adpxr2fmy25jzcoy00htka0awzckl.png)
Now, apply the trigonometric identity:
![\begin{gathered} tan\theta=(sin\theta)/(cos\theta) \\ \\ tan\theta=(0.9629)/(-0.27) \\ \\ tan\theta=−3.57 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/munws1dx86sqlqifwbsmrxf7qisi8ofteg.png)
Since tan is positive in the third quadrant, we have:
![tan\theta=3.57](https://img.qammunity.org/2023/formulas/mathematics/college/7taxqryazgg4ru36mgtlti04vheazqtudd.png)
ANSWER:
3.57