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# Trig Issue - Is this step even allowed? watch

1. Hi there,

here is the question:

The issue I have is this step:

I don't understand how they are able to multiply ONLY the denominator to make it the same when surely you have to multiply the numerator too??
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2. (Original post by CrazyFool229)
Hi there,

here is the question:

The issue I have is this step:

I don't understand how they are able to multiply ONLY the denominator to make it the same when surely you have to multiply the numerator too??
Multiplying the numerator and denominator of 1/1 by cos 2u gives cos 2u / cos 2u.
Thus we have cos 2u / cos 2u + 1 / cos 2u = (cos 2u + 1) / cos 2u, as required.
3. i would factorise out a cosx on the top & bottom then cancel... this leaves ( 1 - cosx ) / ( 1 + cosx )... then look at cosx as being cos( 2*x/2) and work with identities for cos2θ
4. (Original post by HapaxOromenon3)
Multiplying the numerator and denominator of 1/1 by cos 2u gives cos 2u / cos 2u.
Thus we have cos 2u / cos 2u + 1 / cos 2u = (cos 2u + 1) / cos 2u, as required.
Dear god... Thanks a lot

I just realised when I was looking at my working I made a very basic mistake with my working out which made it very confusing to follow. Your explanation cleared it all up. Thanks again!

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