cos{2sin^(-1)(-x)}=0

Solve: cos{2\ sin^(-1)(-x)}=0

Let cos(2sin^(-1)x)=1/9xgt0 holds true for x=m/n , where m and n are co-prime further alpha and beta are root of quadratic eqation mx^2-nx-m+n=0,(alpha,beta) then point (alpha, beta) lies on the line

The equation of the normal to the curve y=(1+x)^(2y)+cos^(2)(sin^(-1)x) at x=0 is :

{ cos (sin ^(-1) x)}^(2) ={sin (cos ^(-1) x)}^(2)

If f(x)={{:(sin(cos^(-1)x)+cos(sin^(-1)x)",",xle0),(sin(cos^(-1)x)-cos(sin^(-1)x)",",xgt0):} then at x = 0

If sin^(-1)x in (0, (pi)/(2)) , then the value of tan((cos^(-1)(sin(cos^(-1)x))+sin^(-1)(cos(sin^(-1)x)))/(2)) is :

If sin^(-1)x in (0, (pi)/(2)) , then the value of tan((cos^(-1)(sin(cos^(-1)x))+sin^(-1)(cos(sin^(-1)x)))/(2)) is :

The value of tan(sin^(-1)(cos(sin^(-1)x)))tan(cos^(-1)(sin(cos^(-1)x)))." where "x in (0, 1) is equal to

sin^(2)(cos^(-1)x)+cos^(2)(sin^(-1)(sqrt(1-x^(2)))) = ________ (0ltxlt1)