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scoray [572]
3 years ago
5

Tanθ= a. 1 - cot^2 θ b. 1/cot θ c. 1 + sec^2 θ

Mathematics
2 answers:
DIA [1.3K]3 years ago
7 0

Answer:

tanθ =  \frac{sinθ}{cosθ} \\ =  \frac{1}{ \frac{cosθ}{sinθ} }  \\  =  \frac{1}{cotθ}

<u>b. 1/cot θ</u>

erma4kov [3.2K]3 years ago
5 0

Answer:

tanθ =  \frac{sinθ}{cosθ} \\ =  \frac{1}{ \frac{cosθ}{sinθ} }  \\  =  \frac{1}{cotθ}

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Eddi Din [679]

After factoring 3x^3-27x=0 the solutions are x=0 , x=3, x= -3

Step-by-step explanation:

We need to solve by factoring:

3x^3-27x=0

Solving:

3x^3-27x=0

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3x(x^2-9)=0

Now using formula a^2-b^2=(a-b)(a+b)

3x((x)^2-(3)^2)=0\\3x(x-3)(x+3)=0

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So, After factoring 3x^3-27x=0 the solutions are x=0 , x=3, x= -3

Keywords: Factorization

Learn more about Factorization at:

  • brainly.com/question/1464739
  • brainly.com/question/2568692
  • brainly.com/question/10771256
  • brainly.com/question/1414350

#learnwithBrainly

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