1 trigonometry Q.

let y = sin^(-1) (x/a)
sin y = x/a ,
d / dx (sin y) = d / dx (x/a)
cos y dy/dx = 1/a
dy / dx = 1/ (a cosy)

since cos y = √ [1 - (sin y)^2 ] = √ [ 1 - (x/a)^2] = 1/a √ (a^2 - x^2)

so, d [ sin^(-1) (x/a) ] /dx = 1 / { a [ 1/a √ (a^2 - x^2) ] }
= 1 / √ (a^2 - x^2)

that's proved.

hope it can help you :)

Let y=sinx
dy/dx=cosx
dy/dx=[1-sin^2(x)]^0.5=(1-y^2)^0.5
Hence,
sin^(-1)y=x
d[sin^(-1)y]/dy=dx/dy=1/(dy/dx)=1/(1-y^2)^0.5
Therefore,
d[sin^(-1)(x/a)]/dx
={d[sin^(-1)(x/a)]/d(x/a)}{d(x/a)/dx}
={1/[1-(x/a)^2]^0.5}{1/a}
=1/(a^2-x^2)^0.5