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Question: If \(\sinh^{- 1}x = c\text{osec}\text{h}^{\text{–}\text{1}} ⥂ y\),then the correct statement is...

If sinh⁡−1x=cosech–1⥂y\sinh^{- 1}x = c\text{osec}\text{h}^{\text{–}\text{1}} ⥂ y,then the correct statement is

A

x=yx = y

B

xy=−1xy = - 1

C

xy=1xy = 1

D

x+y=0x + y = 0

Answer

xy=1xy = 1

Explanation

Solution

Given that, sinh⁡−1x=cosech–1y\sinh^{- 1}x = \text{cosec}\text{h}^{\text{–1}}y or sin⁡h−1x=sin⁡h−1(1y){\sin h}^{- 1}x = {\sin h}^{- 1}\left( \frac{1}{y} \right) or

x=sinh⁡{sinh⁡−1(1y)}x = \sinh\left\{ \sinh^{- 1}\left( \frac{1}{y} \right) \right\} or x=1yx = \frac{1}{y} ⇒ xy=1xy = 1