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anonymous

  • one year ago

Could someone please help me verify the trig. equation using identities? 1+sec^2xsin^2x=sec^2x

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  1. anonymous
    • one year ago
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    you could start by the first part of the equality \[1+\sec^{2}x*sen^{2}x\] you have to remember these identities : \[secx=1/cosx\] \[sen^{2}x+cos^{2}x=1\] replacing \[1+\frac{ 1^{2} }{\cos^{2}x }*sen^{2}x\] \[\frac{ cos^{2}x+sen^{2}x }{\cos^{2}x}\] \[\frac{ 1 }{\cos^{2}x}=sec^{2}x\]

  2. anonymous
    • one year ago
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    Thanks so much! @baad1994 Could you help me with one more? -5tan^2x+sec^2x=1

  3. anonymous
    • one year ago
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    And haha, I actually understood what steps you took. (: So thank you so much!!

  4. anonymous
    • one year ago
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    nice! :) , are you sure that there is a 5 before tan^2x?

  5. anonymous
    • one year ago
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    Oh, sorry. I meant -tan^2x not 5tan^2x

  6. anonymous
    • one year ago
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    oh ok , no problem n.n \[-tg^{2}x+\sec^{2}x=1\] the same, start by the first part \[-tg^{2}x+\sec^{2}x\] Remember these: 1)\[tgx=\frac{ senx }{ cosx }\] 2)\[secx=\frac{ 1 }{ cosx }\] 3)\[sen ^{2}x+\cos ^{2}x = 1 \] \[\cos ^{2}x = 1-sen ^{2}x \] replacing \[-\frac{ senx^{2} }{ cosx^{2} }+\frac{ 1 }{ cosx^{2} }\] \[\frac{ 1-sen^{2}x }{ \cos ^{2}x }\]

  7. anonymous
    • one year ago
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    when you are going to verify trigonometric expressions you usually have to pass all to sen and cos. And remember that sen^2x+cos^2x=1

  8. anonymous
    • one year ago
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    so remember the equivalences: secx=1/cosx cscx=1/senx tgx=senx/cosx ctgx=cosx/senx and operate

  9. anonymous
    • one year ago
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    Thanks so much!! (:

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