How do you write an algebraic expression that is equivalent to #sec(arcsin(x-1))#?

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Answer 1 · 2017-01-09T15:44:05

sec (x - 1)

#arcsin (x - 1) -> arc (x - 1)#
The equivalent expression is
#sec (x - 1)#, or #1/(cos (x - 1))#

Answer 2 · 2017-01-09T16:43:28

#1/sqrt(2x-x^2), (xne0,2).#

Let #arc sin(x-1)=theta#

#:. sintheta=(x-1)#

Now, Reqd. Exp. #=sec(arc sin(x-1)#

#=sectheta#

But, #sintheta=(x-1) rArr costheta=sqrt{1-sin^2theta}#

#=sqrt{1-(x-1)^2}#

#=sqrt(2x-x^2)#

#:. sectheta=1/sqrt(2x-x^2)#

Hence, the reqd. Exp. #=1/sqrt(2x-x^2), (xne0,2).#