Explanation:
Given that
[tex]sinB=\frac{6}{11}[/tex]
Where
[tex]sinB=\frac{opposite}{hypotenuse}=\frac{6}{11}[/tex]
The next step will be to et the value of the adjacent
[tex]adjacent=\sqrt{11^2-6^2}=\sqrt{121-36}=\sqrt{85}[/tex]
To find the exact values of other functions, we will have to use the
[tex]cosB=\frac{adjacent}{hypotenuse}=\frac{\sqrt{85}}{11}[/tex][tex]tanB=\frac{opposite}{adjacent}=\frac{6}{\sqrt{85}}[/tex]
Also
[tex]cscB=\frac{1}{sinB}=\frac{1}{\frac{6}{11}}=\frac{11}{6}[/tex]
Then
[tex]secB=\frac{1}{cosB}=\frac{1}{\frac{\sqrt{85}}{11}}=\frac{11}{\sqrt{85}}[/tex]
And finally
[tex]cotB=\frac{1}{tanB}=\frac{1}{\frac{6}{\sqrt{85}}}=\frac{\sqrt{85}}{6}[/tex]
