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vladimir2022 [97]
3 years ago
11

Prove that :

Mathematics
2 answers:
anyanavicka [17]3 years ago
7 0

Answer:

Question 3sinθ/ (1 - cotθ) + cosθ / (1 - tanθ) =sinθ / (1 - cosθ/sinθ) + cosθ / (1 - sinθ/cosθ) =sinθ/ [(sinθ - cosθ) / sinθ] + cosθ / [(cosθ - sinθ)/cosθ] =sin²θ/ (sinθ - cosθ) + cos²θ/(cosθ - sinθ) =sin²θ/ (sinθ - cosθ) - cos²θ/(sinθ - cosθ) =(sin²θ - cos²θ) / (sinθ - cosθ) =(sinθ + cosθ)(sinθ - cosθ) / (sinθ - cosθ) =sinθ + cosθ :D

Step-by-step explanation:

Dimas [21]3 years ago
6 0

Answer:

<h3>Question 1</h3>
  • sec²θ + cosec²θ =
  • 1/cos²θ + 1/sin²θ =
  • (sin²θ + cos²θ)/(sin²θcos²θ) =
  • 1 / (sin²θcos²θ) =
  • [(sin²θ + cos²θ)/sinθcosθ]² =
  • (sinθ/cosθ + cosθ/sinθ)² =
  • (tanθ + cotθ)²
<h3>Question 2</h3>
  • (1 - tan²θ) / (1 + tan²θ) =
  • (1 - sin²θ/cos²θ) / (1 + sin²θ/cos²θ) =
  • (cos²θ - sin²θ) / (cos²θ + sin²θ) =
  • (cosθ + sinθ)(cosθ - sinθ) / 1 =
  • (cosθ + sinθ)(cosθ - sinθ)
<h3>Question 3</h3>
  • sinθ/ (1 - cotθ) + cosθ / (1 - tanθ) =
  • sinθ / (1 - cosθ/sinθ) + cosθ / (1 - sinθ/cosθ) =
  • sinθ/ [(sinθ - cosθ) / sinθ] + cosθ / [(cosθ - sinθ)/cosθ] =
  • sin²θ/ (sinθ - cosθ) + cos²θ/(cosθ - sinθ) =
  • sin²θ/ (sinθ - cosθ) - cos²θ/(sinθ - cosθ) =
  • (sin²θ - cos²θ) / (sinθ - cosθ) =
  • (sinθ + cosθ)(sinθ - cosθ) / (sinθ - cosθ) =
  • sinθ + cosθ
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Step-by-step explanation:

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