How do you show that #sec(2x) + tan(2x) + 1 = 2/(1 - tanx)#?
1 Answer
First of all,
#1/(cos2x) +( sin2x)/(cos2x) + 1 = 2/(1 - sinx/cosx)#
#(sin2x + 1)/(cos2x) + 1 = 2/((cosx - sinx)/cosx)#
#(sin2x + 1)/(cos2x) + (cos2x)/(cos2x) = 2/((cosx - sinx)/cosx)#
We now establish the double angle identities. I won't go into the proofs for these, but I would just like to emphasize that these formulas are very important.
•
•
#(2sinxcosx + 1 + cos2x)/(cos2x) = (2cosx)/(cosx - sinx)#
#(2sinxcosx + 2cos^2x - 1 + 1)/(cos^2x - sin^2x) = (2cosx)/(cosx- sinx)#
Factor:
#(2cosx(sinx + cosx))/((cosx + sinx)(cosx - sinx)) = (2cosx)/(cosx - sinx)#
#(2cosx)/(cosx + sinx) = (2cosx)/(cosx - sinx)#
Identity proved!
Hopefully this helps!
