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TRIG HW HELP...identities?

someone help me verify (prove) the identity: (cos^2x-sin^2x)/sinxcosx = cotx-tanx

Please show steps, thank you very much! :)

Update:

Thank you very much xXx! you are a lifesaver haha. but looking at your answer im just like duh, i couldve figured that out. ahahah.

2 Answers

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  • 9 years ago
    Favorite Answer

    cos^2x/sinxcosx - sin^2x/sinxcosx=cotx-tanx

    Cosx/sinx - sinx/cosx=cotx - tanx

  • 4 years ago

    ok, it is the way you do it with calculus. enable's say a= first selection b=2nd numbers a + b = 20 a^2+b^2 = some selection now you resolve the 1st equation for a or b, it is not significant a = 20 - b and you replace (20-b)^2+b^2= some selection now you opperate (b^2-40*b+4 hundred)+b^2 = 2*b^2-40*b+4 hundred you could now do 2 issues, you could graph or divide, in case you graph, then graph the function f(b)=2*b^2-40*b+4 hundred now in case you derive you get f'(b)= 4*b-40 and you're making it equatl to 0 4*b-40 = 0 and you resolve for b b = 40/4 or b = 10 and you could now get a a = 20-b a = 20-10 = 10 so then you certainly get that a = 10 and b = 10 that provide you the minimum you go with and you do the comparable with selection 2 a + b = 30 a^2+b^2 = some selection now you resolve the 1st equation for a or b, it is not significant a = 30 - b and you replace (30-b)^2+b^2= some selection now you opperate (b^2-60*b+900)+b^2 = 2*b^2-60*b+900 you could now do 2 issues, you could graph or divide, in case you graph, then graph the function f(b)=2*b^2-60*b+900 now in case you derive you get f'(b)= 4*b-60 and you're making it equivalent to 0 4*b-60 = 0 and you resolve for b b = 60/4 or b = 15 and you could now get a a = 30-b a = 30-15 = 15 so then you certainly get that a = 15 and b = 15 that provide you the minimum you go with desire this permits. pd. equivalent to 0 once you derive through fact meaning that the slope is going to be 0 and the slope turns into 0 whilst this is optimal or minimum values.

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