An identity question.

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An identity question.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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(1-sin x)/(1+sin x) = (sec x - tan x)^2
Working with the right side only. sec x = 1/cosx tan x = sinx/cosx Then you have a common denominator of cosx so you would have (1 - sinx) all over cos x then that is ^2. if you square both the top and the bottom you get (1 - sinx)(1 - sinx) all over cos^2x Do you know the identity sin^2x + cos^2x = 1 (This is the most important identity you should know) Solving this for cos^2x you get cos^2x = 1 - sin^2x... So then you substitute that in for cos^2x on the bottom and you get (1 - sinx)(1-sinx) over 1 - sin^2x But... the bottom is the difference of two squares so the bottom becomes (1 + sinx) (1 - sinx) ... Then you should see that the 1 - sinx cancels on the top and the bottom to leave the right side with 1 - sinx over 1 + sinx
Well to add the first two fractions you need to get the LCD, for these two fractions it would be this: (6)7/(6)9 + (9)1/(9)6

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opps
Thank you very much! It's a relief to know that I got half way through correctly!
Really important to practice these. It is always easy to see when someone else does it... But.. It is important that you can do it by yourself.. Keep working at it.
Thank you, I will. XD

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