The value of `cos(2cos^-1x+sin^-1x) at x= 1/5` is (A) 1 (B) 3 (C) 0 (D) `(-2sqrt(6))/5`

the value of cos(cos^-1x+sin^-1(x-2)) is equal to (A) 0 (B) 1 (C) -1 (D) sqrt(1-x^2).sqrt(x^2-4x+3) +x (x-2)

Find the value of cos( 2 cos^(-1) x + sin^(-1) x) " when " x = 1/5

Find the value of cos(2cos^(-1)x+sin^(-1)x) at x=(1)/(5), where 0<=pi and -(pi)/(2)<=sin^(-1)x<=(pi)/(2)

If x=(1)/(5) , the value of cos(cos^(-1)x+2sin^(-1)x) is :

If sin^(-1)x+sin^(-1)y=(2 pi)/(3) ,then the value of cos^(-1)x+cos^(-1)y is (A) (2 pi)/(3) (B) (pi)/(3) (C) (pi)/(2) (D) pi

Solution set of inequation ( cos^-1x)^2-(sin^-1x)^2gt0 is (A) [0, 1/sqrt(2)) (B) [-1, 1/sqrt(2)) (C) (-1,sqrt2) (D) none of these

The value of sin^(-1)(cos(cos^(-1)(cos x)+sin^(-1)(sin x))) where x in((pi)/(2),pi), is equal to (pi)/(2)(b)-pi(c)pi (d) -(pi)/(2)