Solution:
Given the triangle below:
To find the exact values of tan θ, cos θ, and csc θ.
[tex]\begin{gathered} adj=\sqrt{(hyp)^2-(opp)^2} \\ =\sqrt{8^2-7^2} \\ =\sqrt{15} \end{gathered}[/tex]
A) Tan θ:
[tex]\begin{gathered} tan\theta=\frac{opposite}{adjacent} \\ =\frac{7}{\sqrt{15}}\times\frac{\sqrt{15}}{\sqrt{15}} \\ \Rightarrow\tan\theta=\frac{7\sqrt{15}}{15} \end{gathered}[/tex]
B) cos θ:
[tex]\begin{gathered} \cos\theta=\frac{adjavent}{hypotenuse} \\ =\frac{\sqrt{15}}{8} \\ \Rightarrow\cos\theta=\frac{\sqrt{15}}{8} \end{gathered}[/tex]
C) csc θ:
[tex]\begin{gathered} csc\theta=\frac{1}{\sin\theta}=\frac{1}{\frac{opposite}{hypotenuse}} \\ =\frac{1}{\frac{7}{8}} \\ \\ \Rightarrow csc\theta=\frac{8}{7} \end{gathered}[/tex]
Hence, we have
[tex]\begin{gathered} \tan\theta=\frac{7\sqrt{15}}{15} \\ \\ \cos\theta=\frac{\sqrt{15}}{8} \\ \\ csc\theta=\frac{8}{7} \\ \end{gathered}[/tex]
