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

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

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Answer 1 · 2016-02-27T21:15:23

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

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

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

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