Solving for F(x) = 1 / (sec x) + cos x

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The discussion revolves around simplifying the function F(x) = 1 / (1 + tan^2 x)^(1/2) - cos x for the interval 0 < x < π/2. Participants suggest substituting (1 + tan^2 x)^(1/2) with sec x, leading to the expression simplifying to 1 / sec x + cos x. The simplification ultimately results in 0, as cos x - cos x equals zero. There is also a mention of a similar problem for the interval π < x < 3π/2, indicating a potential pattern in the solutions. The conversation highlights the importance of verification in mathematical simplifications.
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problem: Simplify F(x) = 1 / ( 1+tan^2 x)^.5 - cos x :0<x<pi/2

ok the only thing i can think of to substitute in was (1+tan^2 x)^.5 = sec^2 x ...

so i got the problem down to 1 / (sec^2 x)^.5 + cos and then become stucks

i m thinken i can just say it is 1 / (sec x) + cos... not sure if that will work or not

thanks in advance for any help
 
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You're on the right track...here's a hint: 1\secx = cosx
 
ya but then cos x - cos x = 0...

(i had actually worked it out using that but i didn't think the answer would be zero... esp since the next problem is the same the only differnce is that it is from pi < x < 3 pi / 2)

i guess i just needed a second untainted opion... thanks
 
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