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Prove that :

Question 1 :
[tex]sec^2 \theta +cosec^2\theta = (tan \theta +cot \theta )^2[/tex]

Question 2 :
[tex]\frac{1-tan^2\theta}{1+tan^2\theta} =(cos +sin \theta)(cos \theta -sin \theta)[/tex]

Question 3 :
[tex]\frac{sin \theta}{1-cot \theta}+\frac{cos \theta}{1-tan \theta} =cos \theta +sin \theta[/tex]

Answer :

mhanifa

Answer:

Question 1

  • sec²θ + cosec²θ =
  • 1/cos²θ + 1/sin²θ =
  • (sin²θ + cos²θ)/(sin²θcos²θ) =
  • 1 / (sin²θcos²θ) =
  • [(sin²θ + cos²θ)/sinθcosθ]² =
  • (sinθ/cosθ + cosθ/sinθ)² =
  • (tanθ + cotθ)²

Question 2

  • (1 - tan²θ) / (1 + tan²θ) =
  • (1 - sin²θ/cos²θ) / (1 + sin²θ/cos²θ) =
  • (cos²θ - sin²θ) / (cos²θ + sin²θ) =
  • (cosθ + sinθ)(cosθ - sinθ) / 1 =
  • (cosθ + sinθ)(cosθ - sinθ)

Question 3

  • 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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