homework help 957

If we so desire, we can also modify the right-hand side to match the left-hand side.

We should write ##sinxtanx## in terms of ##sinx## and ##cosx##, using the identity ##color(red)(tanx=sinx/cosx)##:

##sinxtanx=sinx(sinx/cosx)=sin^2x/cosx##

Now, we use the Pythagorean identity, which is ##sin^2x+cos^2x=1##. We can modify this to solve for ##sin^2x##, so: ##color(red)(sin^2x=1-cos^2x)##:

##sin^2x/cosx=(1-cos^2x)/cosx##

Now, just split up the numerator:

##(1-cos^2x)/cosx=1/cosx-cos^2x/cosx=1/cosx-cosx##

Use the reciprocal identity ##color(red)(secx=1/cosx##:

##1/cosx-cosx=secx-cosx##

Solution:

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