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drasti1211

  • 3 years ago

simplify:-sec^2x/cot^2x+1

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  1. Hero
    • 3 years ago
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    \[\space \space \space -\sec^2x \div (\cot^2x + 1)\] \[=-\frac{1}{\cos^2x} \div \left(\frac{\cos^2x}{\sin^2x} + 1\right)\] \[=-\frac{1}{\cos^2x} \div \left(\frac{\cos^2x}{\sin^2x} + \frac{\sin^2x}{\sin^2x}\right)\] \[=-\frac{1}{\cos^2x} \div \left(\frac{1}{\sin^2x} \right)\] \[=-\frac{1}{\cos^2x} \times \left(\frac{\sin^2x}{1} \right)\] \[=-\frac{\sin^2x}{\cos^2x}\] \[=-\tan^2x\]

  2. drasti1211
    • 3 years ago
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    in my txbook the answer is tan^2x

  3. Hero
    • 3 years ago
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    There's a negative in front that cannot be ignored

  4. Hero
    • 3 years ago
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    Are you sure you wrote down the problem correctly?

  5. drasti1211
    • 3 years ago
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    k wait

  6. drasti1211
    • 3 years ago
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    k it was sec not -sec sorry i just wrote it :- instead of writing it :

  7. drasti1211
    • 3 years ago
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    can u plz help me with other one?

  8. drasti1211
    • 3 years ago
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    cotx-cos^3xcscx

  9. Hero
    • 3 years ago
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    \[\cot x - \cos^3x \csc x \\=\frac{\cos x}{\sin x} - \frac{\cos^3x}{\sin x} \\=\frac{\cos x - \cos^3x}{\sin x} \\=\frac{\cos x(1 - \cos^2x)}{\sin x} \\=\frac{\cos x(\sin^2x)}{\sin x} \\=\cos x \sin x \]

  10. drasti1211
    • 3 years ago
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    thank you so much u rock!i wish u were my pre cal teacher

  11. Hero
    • 3 years ago
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    okay

  12. drasti1211
    • 3 years ago
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    wait how did u get cos^3x-cosx=cosx(1-cos^2x) ?

  13. drasti1211
    • 3 years ago
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    can't we just subtract it out?

  14. drasti1211
    • 3 years ago
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    k i get what u did !thanks for all ur help

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