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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.
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is it really\[tan^4(h1)\] because if so this is a silly problem as tan is just sine over cosine. perhaps \[tan^4(h)1\]?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[tan^4(h)1=(tan^21)(tan^2+1)\] 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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