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anonymous

  • 5 years ago

tan^4 h -1 I need to express this in terms of sin h and cos h and simplify. I typed it wrong earlier. Sorry.

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  1. anonymous
    • 5 years ago
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    is it really\[tan^4(h-1)\] because if so this is a silly problem as tan is just sine over cosine. perhaps \[tan^4(h)-1\]?

  2. anonymous
    • 5 years ago
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    I don't know. This is how the teacher typed it on her worksheet. Could you show me how to simplify tan^4(h)-1. Thanks

  3. anonymous
    • 5 years ago
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    \[tan^4(h)-1=(tan^2-1)(tan^2+1)\] and \[tan^2(h)+1=sec^2(h)\] \[(tan^2(h)-1)sec^2(h)=\frac{tan^2(h)-1}{cos^2(h)}\] \[=\frac{\frac{sin^2(h)}{cos^2(h)}-1}{cos^2(h)}\] \[=\frac{sin^2(h)}{cos^4(h)}-\frac{1}{cos^2(h)}\] or if you prefer \[\frac{sin^2(h)-cos^2(h)}{cos^4(h)}\]

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