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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` |
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