while differentiatingone inverse trigonometric function with respect to another trigonometric function ,we take \"x=tan Y\" OR :\"X=COS Y\" OR \"X=SIN Y\"after the substitution of the x we may get \"tan inverse of tan 2y \" for some \"f(x)\"after with out checking any conditions we write \"f(x)\" as \"2y\"why are we not checking the conditions in all cases?????

dear nikhil , i m explaning this with the help of example ...

suppose we have to diffenertiate sin^-1x wrt cos^-1x ....

this means  , methametically this can be written as dsin^-1x/dcos^-1x

now  , let sin^-1x = Z & cos^-1x = Y

  Z = sin^-1x

 dZ/dx = d/dx (sin^-1x) = 1/(1-x²)^1/2            .................1

Y = cos^-1x

dY/dx = -1/(1-x²)^1/2               .....................2

dividing  1 by  2

dZ/dY = -1

now , Z = sin^-1x & Y = cos^-1x so

differentiation of sin^-1x wrt cos^-1x is -1 ...

now see another example ,

differentiate sin^-1x wrt tan^-1x ...

let Z = sin^-1x  & Y = tan^-1x then

dZ/dx = 1/(1-x²)^1/2 &             ............1

dY/dx = 1/1+x²                 ................2

divide 1 by 2

dZ/dY = 1+x²/(1-x²)^1/2 = differentiation of sin^-1x wrt tan^-1x ...