1.

If `theta in [(pi)/(2),3(pi)/(2)] then sin^(-1)(sin theta)` equalsA. `theta`B. `pi - theta`C. `2pi - theta`D. `-pi + theta`

Answer» We have
`theta in [(pi)/(2),(3pi)/(2)]rarr-theta in[-(3pi)/(2),(pi)/(2)]rarr pi -thetha in[-(pi)/(2)-(pi)/(2)]`
Aslo sin`(pi-theta) =sin theta`
`therefore sin^(-1)(sin-theta)=sin^(-1)sin(pi 0 theta)=pi-theta`


Discussion

No Comment Found