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anonymous
 5 years ago
An identity question.
anonymous
 5 years ago
An identity question.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(1sin x)/(1+sin x) = (sec x  tan x)^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Working 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)(1sinx) 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thank you very much! It's a relief to know that I got half way through correctly!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Really 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thank you, I will. XD
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