# Math

cosx+cotX/secX+tanx=cosxcotx

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prove cosx+cotx/secx+tanx =cosxcotx

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I'm only aloud to manipulate one side of the problem and the end result has to match the other side of the equation Problem 1. sinx + cosx + sinx + tanx + cosxcotx = secx + cscx Problem 2. ((sinx + cosx)/(1 + tanx))^2 + ((sinx -

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Im really struggling with these proving identities problems can somebody please show me how to do these? I'm only aloud to manipulate one side of the equation and it has to equal the other side of the equation at the end Problem

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My previous question: Verify that (secx/sinx)*(cotx/cscx)=cscx is an identity. (secx/sinx)*(cotx/cscx) = (secx/cscx)(cotx/sinx) = (sinx/cosx)*cotx*(1/sinx) "The last steps should be obvious" Not to me. I can convert (sinx/cosx) to

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Trigonometric Identities Prove: (tanx + secx -1)/(tanx - secx + 1)= tanx + secx My work so far: (sinx/cosx + 1/cosx + cosx/cosx)/(sinx/cos x - 1/cosx + cosx/cosx)= tanx + cosx (just working on the left side) ((sinx + 1 -

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1/tanx-secx+ 1/tanx+secx=-2tanx so this is what I did: =tanx+secx+tanx-secx =(sinx/cosx)+ (1/cosx)+(sinx/cosx)-(1/cosx) =sinx/cosx+ sinx /cosx= -2tanxI but I know this can't be correct because what I did doesn't end as a

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