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Question

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

sin⁡h−1(12){\sin h}^{- 1}\left( \frac{1}{2} \right) is equal to.

A

tanh⁡−1(5)\tanh^{- 1}(\sqrt{5})

B

tan⁡h−1(15){\tan h}^{- 1}\left( \frac{1}{\sqrt{5}} \right)

C

tanh⁡−1(3)\tanh^{- 1}(\sqrt{3})

D

tan⁡h−1(25){\tan h}^{- 1}\left( \frac{2}{\sqrt{5}} \right)

Answer

tan⁡h−1(15){\tan h}^{- 1}\left( \frac{1}{\sqrt{5}} \right)

Explanation

Solution

sinh⁡−1(12)=tanh⁡−1x1+x2=tan⁡h−1(121+(12)2)\sinh^{- 1}\left( \frac{1}{2} \right) = \tanh^{- 1}\frac{x}{\sqrt{1 + x^{2}}} = {\tan h}^{- 1}\left( \frac{\frac{1}{2}}{\sqrt{1 + \left( \frac{1}{2} \right)^{2}}} \right)

=tanh⁡−1(121+14)=tan⁡−1(1252)=tan⁡h−1(15)= \tanh^{- 1}\left( \frac{\frac{1}{2}}{\sqrt{1 + \frac{1}{4}}} \right) = \tan^{- 1}\left( \frac{\frac{1}{2}}{\frac{\sqrt{5}}{2}} \right) = {\tan h}^{- 1}\left( \frac{1}{\sqrt{5}} \right).