Anónimo
Anónimo preguntado en Ciencias y matemáticasMatemáticas · hace 2 meses

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### 2 respuestas

Calificación
• hace 2 meses

= sin(a).cot(a) → you know that: cot(a) = cos(a)/sin(a)

= sin(a).[cos(a)/sin(a)] → you can simplify by sin(a)

= cos(a)

= csc(a).tan(a) → you know that: csc(a) = 1/sin(a)

= [1/sin(a)].tan(a) → you know that: tan(a) = sin(a)/cos(a)

= [1/sin(a)].[sin(a)/cos(a)]

= sin(a)/[cos(a).sin(a)] → you can simplify by sin(a)

= 1/cos(a) → you know that: 1/cos(a) = sec(a)

= sec(a)

= csc(a).[1 - cos²(a)] → recall: csc(a) = 1/sin(a)

= [1/sin(a)].[1 - cos²(a)] → you know that: cos²(a) + sin²(a) = 1

= [1/sin(a)].[cos²(a) + sin²(a) - cos²(a)]

= [1/sin(a)].[sin²(a)]

= sin²(a)/sin(a)

= sin(a)

= sin(a).[1 + cot²(a)] → recall: cot(a) = cos(a)/sin(a)

= sin(a).[1 + {cos(a)/sin(a)}²]

= sin(a).[1 + {cos²(a)/sin²(a)}]

= sin(a).[{sin²(a)/sin²(a)} + {cos²(a)/sin²(a)}]

= sin(a).[{sin²(a) + cos²(a)}/sin²(a)] → recall: cos²(a) + sin²(a) = 1

= sin(a).[1/sin²(a)]

= sin(a)/sin²(a)

= 1/sin(a) → recall: 1/sin(a) = csc(a)

= csc(a)

= [tan²(a) + 1].cos²(a) → recall: tan(a) = sin(a)/cos(a)

= [{sin(a)/cos(a)}² + 1].cos²(a)

= [{sin²(a)/cos²(a)} + 1].cos²(a)

= [{sin²(a)/cos²(a)} + {cos²(a)/cos²(a)}].cos²(a)

= [{sin²(a) + cos²(a)}/cos²(a)].cos²(a) → recall: cos²(a) + sin²(a) = 1

= [1/cos²(a)].cos²(a)

= cos²(a)/cos²(a)

= 1

= sec²(a).[1 - sin²(a)] → recall: sec(a) = 1/cos(a)

= [1/cos(a)]².[1 - sin²(a)]

= [1/cos²(a)].[1 - sin²(a)]

= [1 - sin²(a)]/cos²(a) → recall: cos²(a) + sin²(a) = 1

= [cos²(a) + sin²(a) - sin²(a)]/cos²(a)

= [cos²(a)]/cos²(a)

= 1

= sec(a).[1 - sin²(a)] → recall: sec(a) = 1/cos(a)

= [1/cos(a)].[1 - sin²(a)]

= [1 - sin²(a)]/cos(a) → recall: cos²(a) + sin²(a) = 1

= [cos²(a) + sin²(a) - sin²(a)]/cos(a)

= [cos²(a)]/cos(a)

= cos(a)

= [sin²(a) + cos²(a)]/tan²(a) → recall: cos²(a) + sin²(a) = 1

= 1/tan²(a) → recall: tan(a) = sin(a)/cos(a)

= 1/[sin(a)/cos(a)]²

= 1/[sin²(a)/cos²(a)]

= cos²(a)/sin²(a)

= [cos(a)/sin(a)]² → recall: cos(a)/sin(a) = cot(a)

= cot²(a)

= [sin²(a) + cos²(a)]/[tan²(a) + 1] → recall: cos²(a) + sin²(a) = 1

= 1/[tan²(a) + 1] → recall: tan(a) = sin(a)/cos(a)

= 1/[{sin(a)/cos(a)}² + 1]

= 1/[{sin²(a)/cos²(a)} + 1]

= 1/[{sin²(a)/cos²(a)} + {cos²(a)/cos²(a)}]

= 1/[{sin²(a) + cos²(a)}/cos²(a)] → recall: cos²(a) + sin²(a) = 1

= 1/[1/cos²(a)]

= cos²(a)

• hace 2 meses

senA cot A=cos A

sen A cos A/ sen A=cos A

cos A=cos A

cscA tanA= secA

1/senA senA/cosA=secA

1/cosA=secA

secA=secA

cscA(1-cos²A)=senA

1/senA sen²A= senA

senA=senA

senA(1+cot²A)=cscA

senA(1+cos²A/sen²A)=cscA

senA((sen²A+cos²A)/sen²A)=cscA

1/senA=cscA

cscA=cscA

(tan²A+1)cos²A=1

(sen²A/cos²A +1)cos²A=1

((sen²A+cos²A)/cos²A)cos²A=1

(1/cos²A)cos²A=1

1=1

sec²A(1-sen²A)=1

(1/cos²A)cos²A=1

1=1

secA(1-sen²A)=cosA

(1/cosA)cos²A=cosA

cosA=cosA

(sen²A+cos²A)/tan²A=cot²A

1/(sen²A/cos²A)=cot²A

cos²A/sen²A=cot²A

cot²A=cot²A

(sen²A+cos²A)/tan²A+1=cos²A

1/((sen²A/cos²A)+1=cos²A

1/((sen²A+cos²A)/(cos²A))=cos²A

1/1/cos²A=cos²A

cos²A=cos²A

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