Find the value of `theta` if `(3+2isintheta)//(1-2isintheta)` is purely real or purely imaginary.

Options

A. `(pi)/3`
B. `(pi)/6`
C. `sin""^(-1)((sqrt(3))/4)`
D. `sin""^(-1)(1/sqrt(3))`

Answer

D

Explanation

Let `z=(2+3 I sin theta )/(1-2 I sin theta ) ` is purely imaginary . Then we have Re (z)=0
Now , consider `z =(2+3 I sin theta)/(1-2i sin theta)`
`=((2+3 is sin theta )(1+2i sin theta))/((1-2i sin theta)(1+2i sin theta))`
`(2+4i sin theta + 3i sin theta +6i^2 sin ^2 theta)/(1^2-(2i sin theta)^2`
`rArr (2+ 7i sin theta - 6 sin ^2 theta)/(1+4 sin ^2 theta)`
`=(2-6 sin ^2 theta )/(1+4 sin ^ 2 theta)+i(7 sin theta)/(1+4 sin^2 theta)`
`therefore (Re(z)=0)`
`therefore (2-6 sin^2 theta)/(1+ 4sin ^2 theta)=0 rArr 2 =6 sin^2 theta`
`rArr sin^2 theta =1/3`
`rArr sin theta = pm 1/(sqrt(3))`
`rArr theta sin^(-1)(pm 1/(sqrt3)= pm sin^(-1)(1/(sqrt 3)))`