Question

$si{n}^{-1}\frac{z-1}{i}$ where z is non-real, can be the angle of triangle if

• A. Re (z) = 1, Im(z ) = 2
• B. Re (z) = 1, 0 < Im(z) ≤ 1
• C. Re (z) + Im(z) = 0
• D. none of these

Answer 1

The correct option is B Re (z) = 1, 0 < Im(z) ≤ 1

Domain of $si{n}^{-1}\left(x\right)is\Rightarrow -1\le x\le 1andxreal.$

Solution: $si{n}^{-1}\frac{1}{i}(z-1)=si{n}^{-1}\left(\frac{1}{i}\right(x-1)+y)$

$x=1$
$0\le y\le 1$
$Re\left(z\right)=1$
$0\le im\left(z\right)\le 1$