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Prove-the-identity-cosh-1-x-sinh-1-1-x-please-help-




Question Number 5874 by sanusihammed last updated on 02/Jun/16
Prove the identity.   cosh^(−1) (x) = sinh^(−1) ((1/x))    please help.
Provetheidentity.cosh1(x)=sinh1(1x)pleasehelp.
Answered by Yozzii last updated on 02/Jun/16
sinh^(−1) u=ln(u+(√(u^2 +1)))  (u∈R)  Let u=(1/x)      (x≠0).  ⇒sinh^(−1) x^(−1) =ln((1/x)+(1/(∣x∣))(√(1+x^2 )))  Let x>0.⇒∣x∣=x.  ⇒sinh^(−1) x^(−1) =ln(((1+(√(1+x^2 )))/x))    coshy=x  e^y +e^(−y) =2x  e^(2y) −2xe^y +1=0  e^y =((2x±(√(4x^2 −4)))/2)  e^y =x±(√(x^2 −1))  y=ln(x±(√(x^2 −1)))  cosh^(−1) x=ln(x±(√(x^2 −1)))  x≥1.  Let x=1⇒cosh^(−1) 1=ln(1±0)=0  But sinh^(−1) 1=ln(((1+(√2))/1))≠0=cosh^(−1) 1
sinh1u=ln(u+u2+1)(uR)Letu=1x(x0).sinh1x1=ln(1x+1x1+x2)Letx>0.⇒∣x∣=x.sinh1x1=ln(1+1+x2x)coshy=xey+ey=2xe2y2xey+1=0ey=2x±4x242ey=x±x21y=ln(x±x21)cosh1x=ln(x±x21)x1.Letx=1cosh11=ln(1±0)=0Butsinh11=ln(1+21)0=cosh11

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