Can you help me verify the identity?

Question

(cos[(pi/2)-x])/(sin[(pi/2)-x])=tanx

Answer 1

cos (π/2 - x) = sin x

sin (π/2 - x) = cos x

sin x / cos x

tan x

QED

Answer 2

Have you been given the following identities for your library yet?

sin θ = cos (π/2 - θ)

cos θ = sin (π/2 - θ)

tan θ = cot (π/2 - θ)

If you have, the problem becomes sin x / cos x = tan x.

If you haven't, it's a little harder because you have to use the subtraction identities.

sin (α - β) = sin(α)cos(β) - sin(β)cos(α)

cos (α - β) = cos(α)cos(β) + sin(α)sin(β)

[cos(π/2)cos(x)+sin(π/2)sin(x)] / [sin(π/2)cos(x)-sin(x)cos(π/2)] = tan(x)

cos(π/2) = 0

sin (π/2) = 1

[0+sin(x)] / [cos(x)-0] = tan(x)

Answer 3

angle subtraction formulas:

cos ( pi/2 - x) = cos(pi/2)cos(x) + sin(pi/2)sin(x) = 0 * cos(x) + 1*sin(x) = sin(x)

sin(pi/2-x) = sin(pi/2)cos(x) + cos(pi/2)sin(x) = 1*cos(x) +0 * cos(x) = cos(x)

sin over cos is tan