Si :tan(x+y)=1/2
tan(3x-y)=3/4
Halla: cot[2(x-y)]
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Desarrollo:
tan(a-b)=(tan(a)-tan(b))/(1+tan(a)tan(b))
entonces:
tan((3x-y)-(x+y))=(tan(3x-y)-tan(x+y))/(1+tan(3x-y)tan(x+y))
tan(3x-y-x-y)=(3/4-1/2)/(1-(3/4)(1/2))
tan(2x-2y)=(1/4)/(1-3/8)
tan(2(x-y))=(1/4)/(5/8)
tan(2(x-y))=2/5
entonces:
cot(2(x-y))=5/2
un gusto =D
tan(a-b)=(tan(a)-tan(b))/(1+tan(a)tan(b))
entonces:
tan((3x-y)-(x+y))=(tan(3x-y)-tan(x+y))/(1+tan(3x-y)tan(x+y))
tan(3x-y-x-y)=(3/4-1/2)/(1-(3/4)(1/2))
tan(2x-2y)=(1/4)/(1-3/8)
tan(2(x-y))=(1/4)/(5/8)
tan(2(x-y))=2/5
entonces:
cot(2(x-y))=5/2
un gusto =D
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