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anonymous
 one year ago
simplify the expression
anonymous
 one year ago
simplify the expression

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ (1\cot(x))^{2} }{ \cot(x) }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@pooja195, @Australopithecus

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.0http://www.sosmath.com/trig/Trig5/trig5/trig5.html Here is a list of trig identities, just use them to simplify it. As for (1cot(x))^2 you can expand it, then cancel out terms

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.0have you tried expanding (1cot(x))^2 ??

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.0show me the work you have done so far

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah it wasnt working out for me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i dont know how to put up work

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.0Expansion rule: (a+b)^2 = (a+b)(a+b) = a*a + a*b + b*a + b*b = a^2 + 2ab + b^2

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.0Just expand the numerator and I will help you from there

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1438196219013:dw

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.0so you have: \[\frac{1^2+\cot(x)*1+\cot^2(x)}{\cot(x)}\] looks good to me simplified you have: \[\frac{1+\cot(x)+\cot^2(x)}{\cot(x)}\] Now next step you need to apply the fraction addition rule: \[\frac{a+b}{c} = \frac{a}{c} + \frac{b}{c}\] Note: you can only apply this rule when the denominator is the same.

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.0apply the rule and show me what you get

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.0oops made a mistake

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.0you messed up and misplaced a negative in your expansion

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0= 1  2 cotx + cot ^ x  cot x divide each term by cot x note 1 / cot x = tan x

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.0remember you have: (1cot(x))^2 so you need to do the following dw:1438196628184:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so would it be cot(x)+2/

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.0Note:  *  = +  * + =  + *  =  + * + = +

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0( a  b)^2 = a^2  2ab + b^2

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.0so writing it all out you have: 1*1 + 1*(cos(x)) + 1*(cos(x)) + (cos(x))*(cos(x))

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.0this isnt something you want to just memorize formulas for you should know the process of solving an expansion

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so would it be sin(x)cos(x)2?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.0So for example: (a + b)(c + d) ^  first step start with a multiply it by c then multiply it by d you get a*c + a*d Now switch to b (a + b)(c + d) ^  second step multiply b by c then multiply b by d then you get: a*c + a*d + b*c + b*d as for the next step once you have your problem split it into separate fractions makes it easier to visualize: \[\frac{1  2\cot(x) + \cot(x)^2}{\cot(x)} = \frac{1}{\cot(x)}  \frac{2\cot(x)}{\cot(x)} + \frac{\cot^2(x)}{\cot(x)}\]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.0By rule: \[a^{1} = a\] \[\frac{1}{a} = a^{1}\] and \[a^{0} = 1\] Thus: \[\frac{a}{a} = a^{1}a^{1} = a^{1 + (1)} = a^{0} = 1\]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.0so if you had: \[\frac{\sin(x)}{\sin^2(x)} = \sin^1(x)*\sin^{2}(x) = \sin^{12}(x) = \sin^{1}(x) = \frac{1}{\sin(x)}\]
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