Question

why does x=sin^-1[tan60degrees] not exist?

Answer 1

what is tan(60)=?

Answer 2

sqt(3)

Answer 3

and isn't sqrt(3)>1?

Answer 4

\(\Large\color{blue}{ \tan(60)=\sqrt{3}}\)
and when you say,
\(\Large\color{blue}{ x=\sin^{-1}(\sqrt{3})}\)
you are saying that,

\(\Large\color{blue}{ \sin(x)=\sqrt{3}}\)

but sine of an angle has to be betwen (and including) 1 and -1.

Answer 5

yeah, i see now

Answer 6

since a hypotenuse is never smaller than the opposite side.

Answer 7

and sin is a ratio of an opposite side (of a triangle) DIVIDED BY hypotenuse (of the triangle)