How do you prove $cscx = sec(pi/2 - x)$ ?

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Answer 1 · 2018-03-29T03:25:09

Since this is an identity,

the relation is true for all values of x

Given:

$cscx=sec(pi/2-x)$

$cscx=1/sinx$

$sec(pi/2-x)=1/cos(pi/2-x)$

Thus,

$1/sinx=1/cos(pi/2-x)$

$cos(pi/2-x)=sinx$

Since this is an identity,

the relation is true for all values of x

Answer 2 · 2018-03-29T03:25:56

See below

Using:
$secx=1/cosx$
$1/sinx=cscx$
$cos(x-y)=cosxcosy+sinxsiny$

Start:
$cscx=sec(pi/2-x)$

$cscx=1/cos(pi/2-x)$

$cscx=1/(cos(pi/2)cosx+sin(pi/2)*sinx)$

$cscx=1/(cancel(0*cosx)+1*sinx)$

$cscx=1/sinx$

$cscx=cscx$

How do you prove cscx = sec(pi/2 - x)cscx = sec(pi/2 - x)?