The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

tan^2(x/2)=(secx+cosx-2)/(secx-cosx)

I'm having a devil of a time trying to prove this with half-angle identities. Any help would be appreciated.

tan^2(x/2)=(secx+cosx-2)/(secx-cosx)

Not asking for the answer, but clues would help.
Original post by falchion-gpx
I'm having a devil of a time trying to prove this with half-angle identities. Any help would be appreciated.

tan^2(x/2)=(secx+cosx-2)/(secx-cosx)

Not asking for the answer, but clues would help.


I would start with the RHS

Multiply, simplify, change to x/2
Reply 2
Avatar for Hasufel
Hasufel
16
use for both sides the identities for 1/2 angles:

tan⁡(x2)≡t\tan(\frac{x}{2})\equiv t

so, sec⁡(x)≡1+t21−t2,cos⁡(x)≡1−t21+t2\displaystyle \sec(x) \equiv \frac{1+t^{2}}{1-t^{2}}, \cos(x) \equiv \frac{1-t^{2}}{1+t^{2}}

substitute and simplify the RHS (and LHS) with these, and you should get =t^2 = tan^{2}(x/2)
(edited 9 years ago)
Reply 3
Avatar for falchion-gpx
falchion-gpx
OP
Thank you for the identities. I'll put these in my list to memorize.
Original post by falchion-gpx
Thank you for the identities. I'll put these in my list to memorize.


Or put sec x as 1/cos x, tidy it up to get (1-cos x)/(1+cos x) then use the usual double angle formulae.

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you