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Question: \({\sin h}^{- 1}({\sin h}^{- 1}\theta)\) is equal to...

sin⁡h−1(sin⁡h−1θ){\sin h}^{- 1}({\sin h}^{- 1}\theta) is equal to

A

iθi\theta

B

θ\theta

C

−iθ- i\theta

D

π+iθ\pi + i\theta

Answer

θ\theta

Explanation

Solution

sinh⁡−1(sinh⁡−1θ)\sinh^{- 1}(\sinh^{- 1}\theta) = −isin⁡(isinh⁡−1θ)- i\sin(i\sinh^{- 1}\theta) =−isin⁡[i(−isin⁡−1(iθ)]- i\sin\lbrack i( - i\sin^{- 1}(i\theta)\rbrack

=−sin⁡[sin⁡−1(iθ)]= - \sin\lbrack\sin^{- 1}(i\theta)\rbrack = −i.iθ=−i2θ=θ- i.i\theta = - i^{2}\theta = \theta