If θ∈π2,3π2, then sin−1⁡(sin⁡θ) equals

If $θ \in [\frac{π}{2}, \frac{3 π}{2}],$ then $\sin^{- 1} (\sin θ)$ equals

A
$θ$
B
$π - θ$
C
$2 π - θ$
D
$- π + θ$

Solution:

We have,

$θ \in [\frac{π}{2}, \frac{3 π}{2}] \Rightarrow - θ \in [- \frac{3 π}{2}, - \frac{π}{2}] \Rightarrow π - θ \in [- \frac{π}{2}, \frac{π}{2}]$

Also, $\sin (π - θ) = \sin θ$

$∴ \sin^{- 1} (\sin θ) = \sin^{- 1} (\sin (π - θ)) = π - θ .$

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