Answer:
[tex]\text{tan}(\theta)=\frac{\sqrt{119}}{5}[/tex]
Step-by-step explanation:
We have been given that Sandra knows the Pythagorean identity [tex]\text{sin}^2(\theta)+\text{cos}^2(\theta)=1[/tex]. She is told that [tex]0\leq \theta\leq \frac{\pi}{2}[/tex] and [tex]\text{cos}(\theta)=\frac{5}{12}[/tex].
First of all, we will find value of sine theta using the given identity.
[tex]\text{sin}^2(\theta)+\text{cos}^2(\theta)=1[/tex]
[tex]\text{sin}^2(\theta)+(\frac{5}{12})^2=1[/tex]
[tex]\text{sin}^2(\theta)+\frac{25}{144}=1[/tex]
[tex]\text{sin}^2(\theta)+\frac{25}{144}-\frac{25}{144}=1-\frac{25}{144}[/tex]
[tex]\text{sin}^2(\theta)=\frac{144-25}{144}[/tex]
[tex]\text{sin}^2(\theta)=\frac{119}{144}[/tex]
[tex]\text{sin}(\theta)=\sqrt{\frac{119}{144}}[/tex]
[tex]\text{sin}(\theta)=\frac{\sqrt{119}}{12}[/tex]
[tex]\text{tan}(\theta)=\frac{\text{sin}(\theta)}{\text{cos}(\theta)}[/tex]
[tex]\text{tan}(\theta)=\frac{\frac{\sqrt{119}}{12}}{\frac{5}{12}}[/tex]
[tex]\text{tan}(\theta)=\frac{\sqrt{119}*12}{5*12}=\frac{\sqrt{119}}{5}[/tex]
Therefore, [tex]\text{tan}(\theta)=\frac{\sqrt{119}}{5}[/tex].