If −ab×cd=1\frac{-a}{b}\times{\frac{c}{d}}=1b−a×dc=1 then, cd=?\frac{c}{d}=?dc=? AMultiplicativeidentityofMultiplicative identity of Multiplicativeidentityof\frac{-a}{b}$$B$$\frac{a}{b}$CReciprocalofReciprocal of Reciprocalof\frac{-a}{b}$$D−ab\frac{-a}{b}b−aAnswer: ReciprocalofReciprocal of Reciprocalof\frac{-a}{b}$$ Read Explanation: Given:−ab×cd=1\frac{-a}{b}\times{\frac{c}{d}}=1b−a×dc=1Formula:Reciprocal of x = 1x\frac{1}{x}x1Calculation:[−ab]×cd=1[\frac{-a}{b}]\times{\frac{c}{d}}=1[b−a]×dc=1,cd=1×−ba\frac{c}{d}=1\times{\frac{-b}{a}}dc=1×a−bcd=−ba\frac{c}{d}=\frac{-b}{a}dc=a−bReciprocal of (−ab)=(−ba)(\frac{-a}{b})=(\frac{-b}{a})(b−a)=(a−b)∴ cd=(−ba)\frac{c}{d}=(\frac{-b}{a})dc=(a−b)=Reciprocal of (−ab)(\frac{-a}{b})(b−a) Read more in App