How do you solve #tan(x/2)=cscx-cotx#?

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Answer 1 · 2016-09-24T05:01:20

See the Proof given in Explanation Section.

#"The R.H.S.="cscx-cotx#

#=1/sinx-cosx/sinx#

#=(1-cosx)/sinx#

Here, we use the following Identities :

#1-cosx=2sin^2(x/2), and, sinx=2sin(x/2)cos(x/2)#. Hence,

#"The R.H.S.="(2sin^2(x/2))/(2sin(x/2)cos(x/2))#

#=sin(x/2)/cos(x/2)#

#=tan(x/2)#

#"=The L.H.S."#