If `x in (0,pi/2),t h e ns howt h a t` `d/(dx)cos^(-1){7/2(1+cos2x)+(sqrt(sin^2x-48cos^2x))sinx}` `=1+(7sinx)/sqrt(sin^2-48cos^2x)`

If x in(0,(pi)/(2)), then show that cos^(-1)((7)/(2)(1+cos2x)+sqrt((sin^(2)x-48cos^(2)x))sin x)=x-cos^(-1)(7cos x)

If x in (0, (pi)/(2)) , then show that cos^(-1) ((7)/(2) (1 + cos 2 x) + sqrt((sin^(2) x - 48 cos^(2) x)) sin x) = x - cos^(-1) (7 cos x)

d/(dx)[cos(1-x^2)^2]=

(d)/(dx)(sin{2cos^(-1)(sinx)}]=

(d)/(dx)[(1+cos2x+sin2x)/(1+sin2x-cos2x)]=

d/(dx)[sin {2 cos^(-1) (sin x)}] =

(d)/(dx)[sin^(-1){cos(x^(2)-2)}]=

(d)/(dx)(sin^(2)x+cos^(2)x) =

(d)/(dx)[cos^(-1)sqrt(((1+x))/(2))]

[(d)/(dx)[log sqrt((1-cos2x)/(1+cos2x))]