Question

What is `Cot [ arcsin( sqrt5 / 6) ]`?

Answer 1

`sqrt(155)/5`

Explanation:

Start by letting `arcsin(sqrt(5)/6)` to be a certain angle `alpha`

It follows that `alpha=arcsin(sqrt5/6)`
and so
`sin(alpha)=sqrt5/6`

This means that we are now looking for `cot(alpha)`

Recall that : `cot(alpha)=1/tan(alpha)=1/(sin(alpha)/cos(alpha))=cos(alpha)/sin(alpha)`

Now, use the identity `cos^2(alpha)+sin^2(alpha)=1` to obtain `cos(alpha)=sqrt((1-sin^2(alpha)))`

`=>cot(alpha)=cos(alpha)/sin(alpha)=sqrt((1-sin^2(alpha)))/sin(alpha)=sqrt((1-sin^2(alpha))/sin^2(alpha))=sqrt(1/sin^2(alpha)-1)`

Next, substitute `sin(alpha)=sqrt5/6` inside `cot(alpha)`

`=>cot(alpha)=sqrt(1/(sqrt5/6)^2-1)=sqrt(36/5-1)=sqrt(31/5)=color(blue)(sqrt(155)/5)`