Question

What is `Cot(arcsin (-5/13))`?

Answer 1

`cot(arcsin(-5/13))= -12/5`

Explanation:

Let`" "theta=arcsin(-5/13)`

This means that we are now looking for `color(red)cottheta!`

`=>sin(theta)=-5/13`

Use the identity,

`cos^2theta+sin^2theta=1`

**NB : ** `sintheta` is negative so `theta` is also negative.

We shall the importance of this info later.

`=>(cos^2theta+sin^2theta)/sin^2theta=1/sin^2theta`

`=>cos^2theta/sin^2theta+1=1/sin^2theta`

`=>cot^2theta+1=1/sin^2theta`

`=>cot^2theta=1/sin^2x-1`

`=> cottheta=+-sqrt(1/sin^2(theta)-1)`

`=>cottheta=+-sqrt(1/(-5/13)^2-1)=+-sqrt(169/25-1)=+-sqrt(144/25)=+-12/5`

WE saw the evidence previously that `theta` should be negative only.

And since `cottheta` is odd `=>cott(-A)=-cot(A)` Where `A` is a positive angle.

So, it becomes clear that `cottheta=color(blue)+12/5`

REMEMBER what we called `theta` was actually `arcsin(-15/13)`

`=>cot(arcsin(-5/13)) = color(blue)(12/5)`