SOLUTION: verifying trigonometric identities. (2-sin^2x/cosx)=secx+cosx

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Question 592917: verifying trigonometric identities.
(2-sin^2x/cosx)=secx+cosx

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
These identities can be hard because there is no "recipe" that one can memorize and then apply to each one.

When I looked at your problem here is what I saw:
  • The right side is expressed in terms of sec(x) and cos(x)
  • Since sec(x) is the reciprocal of cos(x) it might be helpful to find a way to express the left side in terms of cos(x)
  • The left side is expressed in terms of sin^2(x) and cos(x). It is very easy to replace the sin^2(x) using sin^2(x) + cos^2(x) = 1 or sin^2(x) = 1 - cos^2(x)
  • The right side is two terms while the left side is just one. So at some point we will need to find a way to split the one term into two.

So let's put these ideas into action:
%282-sin%5E2%28x%29%29%2Fcos%28x%29+=+sec%28x%29+%2B+cos%28x%29
Replacing sin^2(x) with 1-cos^2(x) we get:
%282-%281-cos%5E2%28x%29%29%29%2Fcos%28x%29+=+sec%28x%29+%2B+cos%28x%29
Note the use of parentheses! This is an extremely good habit (in any expression, not just Trig expressions) whenever making substitutions involving different numbers of terms. (Here we are replacing 1 term with two.)

Simplifying the numerator carefully we get:
%281%2Bcos%5E2%28x%29%29%2Fcos%28x%29+=+sec%28x%29+%2B+cos%28x%29
(I hope it is clear why there is a "+" in front of the cos^2(x) and how it would be very easy to get this wrong without the parentheses!)

Now we will split the fraction into to terms (by "un-adding" them):
1%2Fcos%28x%29%2Bcos%5E2%28x%29%2Fcos%28x%29+=+sec%28x%29+%2B+cos%28x%29
The first fraction is sec(x) and we can cancel a factor of cos(x) in the second fraction leaving us with:
sec(x) + cos(x) = sec(x) + cos(x)
And we're done!