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Question: \(2{\sin h}^{- 1}(\theta)\) is equal to...

2sin⁡h−1(θ)2{\sin h}^{- 1}(\theta) is equal to

A

sinh⁡−1(2θ1+θ2)\sinh^{- 1}(2\theta\sqrt{1 + \theta^{2}})

B

sin⁡h−1(2θ1−θ2){\sin h}^{- 1}(2\theta\sqrt{1 - \theta^{2})}

C

sin⁡h−1(θ1+θ2){\sin h}^{- 1}(\theta\sqrt{1 + \theta^{2}})

D

None of these

Answer

sinh⁡−1(2θ1+θ2)\sinh^{- 1}(2\theta\sqrt{1 + \theta^{2}})

Explanation

Solution

We know that, 2sin⁡−1x=sin⁡−1(2x1−x2)2\sin^{- 1}x = \sin^{- 1}(2x\sqrt{1 - x^{2}})

Putting the value of x=iθx = i\theta

2sin⁡−1(iθ)=sin⁡−1(2iθ1−i2θ2)2\sin^{- 1}(i\theta) = \sin^{- 1}(2i\theta\sqrt{1 - i^{2}\theta^{2})}

2isin⁡h−1(θ)=sin⁡−1(2iθ1+θ2)2i{\sin h}^{- 1}(\theta) = \sin^{- 1}(2i\theta\sqrt{1 + \theta^{2})} or

2isinh⁡−1(θ)=isinh⁡−1(2θ1+θ2)2i\sinh^{- 1}(\theta) = i\sinh^{- 1}(2\theta\sqrt{1 + \theta^{2}}) (∵sin⁡−1(ix)=isin⁡h−1x\because\sin^{- 1}(ix) = i{\sin h}^{- 1}x)

⇒ 2sin⁡h−1(θ)=sin⁡h−1(2θ1+θ2)2{\sin h}^{- 1}(\theta) = {\sin h}^{- 1}(2\theta\sqrt{1 + \theta^{2}})