Answer:
[tex]\textsf{D.}\quad\dfrac{b}{\sqrt{a^2+b^2}}[/tex]
Step-by-step explanation:
Tan trig ratio
[tex]\tan(x)=\sf \dfrac{O}{A}[/tex]
where:
- [tex]x[/tex] is the angle
- O is the side opposite the angle
- A is the side adjacent the angle
Using the given information:
[tex]\implies x=\tan^{-1}\left(\dfrac{a}{b}\right)[/tex]
[tex]\implies \tan(x)=\tan\left(\tan^{-1}\left(\dfrac{a}{b}\right)\right)[/tex]
[tex]\implies \tan(x)=\left(\dfrac{a}{b}\right)[/tex]
Comparing this with the trig ratio, we can say that:
[tex]x[/tex] is the angle opposite side [tex]a[/tex]
Cos trig ratio
[tex]\cos(x)=\sf \dfrac{A}{H}[/tex]
where:
- [tex]x[/tex] is the angle
- A is the side adjacent the angle
- H is the hypotenuse
Therefore:
- A = side [tex]b[/tex]
- H = side [tex]\sqrt{a^2+b^2}[/tex]
[tex]\implies \cos\left(\tan^{-1}\left(\dfrac{a}{b}\right)\right)=\cos(x)= \dfrac{b}{\sqrt{a^2+b^2}}[/tex]
