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anonymous

  • one year ago

How do I substitute identities to make the right side match the left? cotx sec^4x = cotx+2tanx+tan^3x (sinx)(tanx cosx-cotx cosx) = 1-2cos^2x 1+sec^2x sin^2x = sec^2x (sinx/1-cosx)+(sinx/1+cosx) = 2cscx

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  1. 0_0youscareme
    • one year ago
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    1. The key to this one is that sec^2(x) = 1 + tan^2(x). So the left side is cot(x) (1 + tan^2(x)) (1 + tan^2(x)) Expanding this term by term you get cot(x) + 2 cot(x) tan^2(x) + cot(x) tan^4(x), but since cot(x) tan(x) = 1, that turns into cot(x) + 2 tan(x) + tan^3(x), which is the same as the right side. #2. Just copying it first... (sin x)(tan x cos x - cot x cos x) = 1 - 2 cos^2 x (sin x)[ (sin x/cos x)(cos x) - (cos x/sin x)(cos x) ] = (1 - cos^2 x) - cos^2 x (sin x) [ sin x - (cos^2 x/sin x) ] = sin^2 x - cos^2 x sin^2 x - cos^2 x = sin^2 x - cos^2 x #3. 1 + sec^2(x) sin^2(x) = sec^2(x) But sec(x) = 1/cos(x), so sec(x)sin(x) = tan(x), so the left side becomes 1 + tan^2(x) = sec^2(x) This is a well-known Pythagorean identity. #4. Ugh! ...OK ... (sin x)/(1 - cos x) + (sin x)/(1 + cos x) = 2 csc x Multiply all three terms by (1 - cos x)(1 + cos x), obtaining (sin x)(1+cos x) + (sin x)(1 - cos x) = 2 csc x (1 - cos^2 x) But 1 - cos^2 x is sin^2 x, so this becomes (sin x)(1+cos x) + (sin x)(1 - cos x) = 2 csc x (sin^2 x) sin x + sin x cos x + sin x - sin x cos x = 2 sin x 2 sin x = 2 sin x #5. The right-hand side is just a 1, since sec x = 1 / cos x. We have - tan^2 x + sec^2 x = 1, or sec^2 x = 1 + tan^2 x, again the well-known Pythagorean identity.

  2. alekos
    • one year ago
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    you're not supposed to spoonfeed the answers

  3. alekos
    • one year ago
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    get'em to do a bit of work

  4. 0_0youscareme
    • one year ago
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    sorry

  5. anonymous
    • one year ago
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    Thank you for your help. I really appreciate the step by step, it helped me to better see what I have to do. Next time though, instead of giving me the answer please walk me or anyone else for that matter through it. It will benefit the person more than just giving them the answers. But yet again, I appreciate you doing that for me, thank you.

  6. alekos
    • one year ago
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    i'm glad you see it that way

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