How do you verify the identity #cosxcotx = cscx - sinx#?

2 Answers
Mar 17, 2018

All the identities you will need:
#cotx= cosx/sinx#
#cos^2x= 1-sin^2x#
#1/sinx= cscx#

Starting:
#cosxcotx= cscx-sinx#

Apply number 1 on the list:

#cosx*cosx/sinx= cscx-sinx#

Simplify:

#cos^2x/sinx= cscx-sinx#

Apply number 2 on the list:

#(1-sin^2x)/sinx= cscx-sinx#

Split the numerator:

#1/sinx-sin^2x/sinx= cscx-sinx#

Apply number 3 on the list:

#cscx-sinx=cscx-sinx#

Mar 17, 2018

Kindly go through a Proof in the Explanation.

Explanation:

We have, #cosxcotx+sinx#,

#=cosx*cosx/sinx+sinx#,

#=(cos^2x+sin^2x)/sinx#,

#=1/sinx#,

#:. cosxcotx+sinx=cscx#.

#rArr cosxcotx=cscx-sinx#, as desired!