Question #5cf65

1 Answer
Nov 27, 2016

Currently, as is, the identity is false. If we change #sinx - cosx# in the denominator on the left to #sinx + cosx#, it will be true though (thanks to Mrs. R. for pointing this out).

We know that #tantheta = sintheta/costheta#, #cottheta = 1/tantheta = 1/(sintheta/costheta) = costheta/sintheta#, #sectheta = 1/costheta# and #csctheta = 1/sintheta#. So:

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

#((cos^2x - sin^2x)/(sinxcosx))/(sinx + cosx) = (cosx - sinx)/(cosxsinx)#

#((cosx + sinx)(cosx - sinx))/((sinxcosx)(cosx + sinx)) =(cosx - sinx)/(cosxsinx)#

#(cosx - sinx)/(sinxcosx) = (cosx - sinx)/(cosxsinx)#

#LHS = RHS#

Identity proved!

Hopefully this helps!