Prove that : sin^(-1)x+cos^(-1)x= π/2 [π is in radian]

sin^(-1)x+cos^(-1)x= π/2


Let, sin^(-1)x=θ; then sinθ= x



now, sinθ = cos( π/2 - θ)=x and cos^(-1)x= π/2 - θ

therefore, sin^(-1)x+cos^(-1)x= θ+π/2 - θ= π/2


Hence, Proved.

Remember these identities:

sin(a + b) = sin(a)cos(b) + sin(b)cos(a)
sin(x)^2 + cos(x)^2 = 1


arcsin(x) + arccos(x) = pi/2
sin(arcsin(x) + arccos(x)) = sin(pi/2)
sin(arcsin(x)) * cos(arccos(x)) + sin(arccos(x)) * cos(arcsin(x)) = 1
x * x + sin(arccos(x)) * cos(arcsin(x)) = 1

sin(arccos(x))^2 + cos(arccos(x))^2 = 1
sin(arccos(x))^2 + x^2 = 1
sin(arccos(x))^2 = 1 - x^2
sin(arccos(x)) = sqrt(1 - x^2)

sin(arcsin(x))^2 + cos(arcsin(x))^2 = 1
x^2 + cos(arcsin(x))^2 = 1
cos(arcsin(x)) = sqrt(1 - x^2)

x^2 + sqrt(1 - x^2) * sqrt(1 - x^2) = 1
x^2 + (1 - x^2) = 1
x^2 + 1 - x^2 = 1
1 = 1

Proven!

take the derivative of the left side to show it is 0 ---> left side is constant , then choose x = 0

to find the constant