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Question Number 85386 by john santu last updated on 21/Mar/20
If cot x−cos x = n and m^2 −n^(2 ) = 4(√(mn)) prove that m = cot x + cos x
Ifcotx−cosx=nandm2−n2=4mnprovethatm=cotx+cosx
Answered by mind is power last updated on 23/Mar/20
n=a−b m=(a+b) ⇒m^2 −n^2 =4ab=4(√(a^2 −b^2 )) ⇒a^2 b^2 =a^2 −b^2 ⇒a^2 =(b^2 /(1−b^2 )) ⇒n=(b/( (√(1−b^2 ))))−b⇒=cot(x)−cos(x) f(b)=(b/( (√(1−b^2 ))))−b f′(b)=(((√(1−b^2 ))+(b^2 /( (√(1−b^2 )))))/((1−b^2 )))−1 =(1/((1−b^2 )(√(1−b^2 ))))−1>0 since b∈]−1,1[ f(b) is increase function f(cos(x))=cot(x)−cos(x)⇒b=cos(x)⇒ a=cot(x)⇒m=a+b=cot(x)+cos(x)
n=a−bm=(a+b)⇒m2−n2=4ab=4a2−b2⇒a2b2=a2−b2⇒a2=b21−b2⇒n=b1−b2−b⇒=cot(x)−cos(x)f(b)=b1−b2−bf′(b)=1−b2+b21−b2(1−b2)−1=1(1−b2)1−b2−1>0sinceb∈]−1,1[f(b)isincreasefunctionf(cos(x))=cot(x)−cos(x)⇒b=cos(x)⇒a=cot(x)⇒m=a+b=cot(x)+cos(x)

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