How do you simplify #Sinx/(1-secx) - sin/(1+secx)#?

1 Answer
Apr 19, 2016

We must remember that #sec = 1/cos#

Explanation:

#sinx/(1 - 1/cosx) - sinx/(1 + 1/cosx)#

#sinx/((cosx - 1)/cosx) - sinx/((cosx + 1)/cosx)#

#sinx xx cosx/(cosx - 1) - sinx xx cosx/(cosx + 1)#

#(sinxcosx)/(cosx - 1) - (sinxcosx)/(cosx + 1)#

#((sinxcosx)(cosx + 1))/((cosx - 1)(cosx + 1)) - ((sinxcosx)(cosx- 1))/((cosx + 1)(cosx - 1))#

#(sinxcos^2x + sinxcosx)/(cos^2x - 1) - (sinxcos^2x - sinxcosx)/(cos^2x - 1)#

#(sinxcos^2x + sinxcosx - sinxcos^2x + sinxcosx)/(cos^2x - 1)#

#(2sinxcosx)/(cos^2x - 1)#

Applying the double angle identity #2sinxcosx = sin2x# and the pythagorean identity #cos^2x - 1 = -sin^2x# we get the following:

#(sin2x)/(-sin^2x) -># this is your answer.

*Beware: This is not the only possible answer; there are many ways of attacking this problem, with different answers completely within the realm of possibility. *

Hopefully this helps!