Answer:
cos∅=-[tex]\frac{\sqrt{85} }{11}[/tex]
tan∅ -[tex]-\frac{6\sqrt{85} }{85}[/tex]
Step-by-step explanation:
sin∅ =[tex]\frac{6}{11}[/tex]
we know that
cos∅ =[tex]\sqrt{1-sin^2∅ }[/tex]
plugging the value of sin∅ gives
[tex]\sqrt{ 1-(\frac{6}{11})^2\sqrt{ 1-(\frac{36}{121})\\\sqrt{\frac{85}{121} } \\\frac{\sqrt{85} }{11\\}[/tex]
therefore cos∅=-[tex]\frac{\sqrt{85} }{11}[/tex]
cos∅ is negative since sec∅ is negative
tan∅ = [tex]\frac{sin∅}{cos∅}[/tex]
=[tex]\frac{\frac{6}{11} }{\frac{\-sqrt{85} }{11} } \\[/tex]
on simplifying it ,we get
=-[tex]\frac{6}{\sqrt{85} }[/tex]
rationalizing it we get
-[tex]-\frac{6\sqrt{85} }{85}[/tex]
