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Answer 1 · 2017-11-17T12:57:50

#tan(sin^-1(x-1))= (x-1)/sqrt(x(2-x)#

Let #sin^-1(x-1)=theta :. sin theta= (x-1) #

In reference right triangle #sin theta = p/h= (x-1)/1=(x-1)#

#:. p=(x-1) ; h=1 :. b= sqrt(1^2-(x-1)^2#

or #b = sqrt( 1-(x^2-2x+1)) = sqrt(2x-x^2) =sqrt(x(2-x)#

#tan theta= p/b = (x-1)/sqrt(x(2-x)#

[p=perpendicular , b=base , h= hypotenuse of rigt triangle]

#tan(sin^-1(x-1))= tan theta= (x-1)/sqrt(x(2-x)# [Ans]