Question

Evaluate sin(Tan^-10)

Answer 1

`\tan x=0` for
`x=n\pi`, where
`n` is any integer. The inverse tangent function returns numbers between
`-\frac\pi2` and
`\frac\pi2`. The only multiple of
`\pi` in this range is 0, so
`\tan^(-1)0=0`.

Then

`\sin\left(\tan^(-1)0\right)=\sin0=\boxed0`