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,

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

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

$=tan(x/2)$

$"=The L.H.S."$

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