Question #0b5d8

1 Answer
ProfLayton
Dec 5, 2017

Proof below

Explanation:

First know these three identities:
#sin2x=2sinxcosx#
#cos2x=cos^2x-sin^2x#
#sin^2x+cos^2x=1#

First, we change the #sin2x# and #cos2x#
#(1-sin2x)/(cos2x)=(1-2sinxcosx)/(cos^2x-sin^2x)#

Now this part may be hard to see, but you need to split the #1# in the numerator using the third identity
#=(cos^2x+sin^2x-2sinxcosx)/(cos^2x-sin^2x)#

Rearrange to look more like the expansion of a square
#=(cos^2x-2sinxcosx+sin^2x)/(cos^2x-sin^2x)#

Factorise the numerator, while using the difference two swaures the factor the denominator
#=(cosx-sinx)^2/((cosx-sinx)(cosx+sinx))#

Cancel like terms:
#=(cosx-sinx)/(cosx+sinx)#

And we're done

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