Let `z=(costheta+isintheta)/(costheta-isintheta), pi/4 lt theta lt pi/2`. Then arg(z) =

Options

A. `2theta`
B. `2theta-pi`
C. `pi+2theta`
D. none of these

Answer

A

Explanation

We have,
`z=(costheta+isintheta)/(costheta-isintheta)= (costheta+isintheta)/(costheta-isintheta) xx (costheta + isintheta)/(costheta+isintheta)`
`rArr z=(costheta+isintheta)^(2)/(cos^(2)theta+sin^(2)theta)=cos2theta+isin2theta`
Now, `pi/4 lt theta lt pi/2`
`rArr pi/2 lt 2theta lt pi`
`rArr cos2theta lt 0` and `sin 2theta gt 0`
`rArr (cos2theta, sin2theta)` is in the second quadrant,
Let `tanalpha=|(("Im(z)"))/("Re(z)")|`. Then,
`rArr tanalpha=|tan2theta|=-tan2theta` `[therefore tan 2theta lt 0]`
`rArr tanalpha=tan(pi-2theta)`
`rArr alpha=pi-2theta`
Since, z is represented by a point in the second quadrant.
`therefore` arg(z) `=pi-alpha=pi-(pi-2theta)=2theta`