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Trig Identities

Prove

sec2x = 1 + tan2x

I did R.H.S

= 1 + sin2x / cos2x
= 1 + (1 - cos2x) / (1 - sin2x)
= 1 + cot2x
= ???

I think I'm doing it right, but I don't know how to solve now.
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2 Answers

Hi Lawson,

Lets go back to the second line of what you have done so far:

= 1 + (1- cos2x)/(1 - sin2x)
 
We can continue from above by looking at the equation as the addition of a whole number to a fraction. The whole number is the 1 and the fraction is (1- cos2x)/(1 - sin2x). Therefore we first need to convert the whole number to a fraction with the same denominator:
 
=> 1 + (1- cos2x)/(1 - sin2x)
= (1 - sin2x)/(1 - sin2x) + (1- cos2x)/(1 - sin2x) = [(1 - sin2x) + (1 - cos2x)]/ (1 - sin2x)
 
Now if we replace all the 1s with cos2 + sin2x, we will have:
 
[(cos2x + sin2x - sin2x) + (cos2x + sin2x - cos2x)]/ (cos2x + sin2x - sin2x)
= [cos2x + sin2x]/cos2x
= 1/cos2x
= sec2x
 
Hope this helps!
A better approach working with the RHS is to  write 
 
1 + sin2x/cos2x  =        (cos2x + sin2x)/cos2x  =  1/ cos2x  =  sec2x
 
The first step is essentially adding fractions, the second used the Pythagorean identity.