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#