How do you prove $\frac{cos [(\frac{π}{2}) - x]}{sin [(\frac{π}{2}) - x]} = tan x$ $cos[(pi/2)-x]/sin[(pi/2)-x]=tanx$ ?

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Answer 1 · 2015-11-01T10:59:36

It's because $cos (\frac{π}{2} - x) = sin (x)$ $cos(pi/2 -x) = sin(x)$ , and $sin (\frac{π}{2} - x) = cos (x)$ $sin(pi/2 -x) = cos(x)$

You need to use two simple trigonometric equality, namely

${ (cos(pi/2 -x) = sin(x)), (sin(pi/2 -x) = cos(x)):}$

Using these two equalities, your expression becomes

$sin(x)/cos(x)$ , which is exactly the definition of $tan(x)$

How do you prove cos[(pi/2)-x]/sin[(pi/2)-x]=tanxcos[(π2)−x]sin[(π2)−x]=tanxcos[(pi/2)-x]/sin[(pi/2)-x]=tanx?