In a right triangle, angle is an acute angle and sin angle = 4/9. Evaluate the other five trigonometric functions of angle.

Answer:Part 1) [tex]cos(A)=\frac{\sqrt{65}}{9}[/tex]Part 2) [tex]tan(A)=4\frac{\sqrt{65}}{65}[/tex]Part 3) [tex]cot(A)=\frac{\sqrt{65}}{4}[/tex]Part 4) [tex]sec(A)=9\frac{\sqrt{65}}{65}[/tex]Part 5) [tex]csc(A)=\frac{9}{4}[/tex]Step-by-step explanation:LetA------> the anglewe have that[tex]sin(A)=\frac{4}{9}[/tex]step 1Find the cos(A)we know that[tex]cos^{2}(A)+sin^{2}(A)=1[/tex]substitute the value of sin(A)[tex]cos^{2}(A)+(\frac{4}{9})^{2}=1[/tex][tex]cos^{2}(A)=1-(\frac{4}{9})^{2}[/tex][tex]cos^{2}(A)=1-(\frac{16}{81})[/tex][tex]cos^{2}(A)=(\frac{65}{81})[/tex][tex]cos(A)=\frac{\sqrt{65}}{9}[/tex]step 2Find the tan(A)we know that[tex]tan(A)=\frac{sin(A)}{cos(A)}[/tex]substitute the values[tex]tan(A)=\frac{\frac{4}{9}}{\frac{\sqrt{65}}{9}}[/tex][tex]tan(A)=\frac{4}{\sqrt{65}}[/tex][tex]tan(A)=4\frac{\sqrt{65}}{65}[/tex]step 3Find the cot(A)we know that[tex]cot(A)=\frac{1}{tan(A)}[/tex]we have[tex]tan(A)=\frac{4}{\sqrt{65}}[/tex]substitute[tex]cot(A)=\frac{1}{\frac{4}{\sqrt{65}}}[/tex][tex]cot(A)=\frac{\sqrt{65}}{4}[/tex]step 4Find the sec(A)we know that[tex]sec(A)=\frac{1}{cos(A)}[/tex]we have[tex]cos(A)=\frac{\sqrt{65}}{9}[/tex]substitute[tex]sec(A)=\frac{1}{\frac{\sqrt{65}}{9}}[/tex][tex]sec(A)=\frac{9}{\sqrt{65}}[/tex][tex]sec(A)=9\frac{\sqrt{65}}{65}[/tex]step 5Find the csc(A)we know that[tex]csc(A)=\frac{1}{sin(A)}[/tex]we have[tex]sin(A)=\frac{4}{9}[/tex]substitute the value[tex]csc(A)=\frac{1}{\frac{4}{9}}[/tex][tex]csc(A)=\frac{9}{4}[/tex]