Question

How do you simplify `Cos[(pi/2)-x]/sin[(pi/2)-x]`?

Answer 1

tanx

Explanation:

Expand numerator and denominator using appropriate `color(blue) " Addition formulae "`

`• cos(A ± B ) = cosAcosB ∓ sinAsinB`

`• sin(A ± B ) = sinAcosB ± cosAsinB`

`color(red) " Numerator "`
`cos( pi/2 - x ) = cos(pi/2)cosx + sin(pi/2)sinx`

now `cos(pi/2) = 0 " and " sin(pi/2) = 1`

simplifies to : 0 + sinx = sinx

`color(orange) " Denominator "`

`sin(pi/2 - x ) = sin(pi/2)cosx + cos(pi/2)sinx`

simplifies to : cosx + 0 = cosx

`rArr cos(pi/2 -x )/sin(pi/2 -x) = sinx/cosx = tanx`