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1 vote
Prove that :

Question 1 :

sec^2 \theta +cosec^2\theta = (tan \theta +cot \theta )^2

Question 2 :

(1-tan^2\theta)/(1+tan^2\theta) =(cos +sin \theta)(cos \theta -sin \theta)

Question 3 :

(sin \theta)/(1-cot \theta)+(cos \theta)/(1-tan \theta) =cos \theta +sin \theta

2 Answers

3 votes

Please find attached herewith the solutions of your questions.

Hope it helps.

Prove that : Question 1 : sec^2 \theta +cosec^2\theta = (tan \theta +cot \theta )^2 Question-example-1
Prove that : Question 1 : sec^2 \theta +cosec^2\theta = (tan \theta +cot \theta )^2 Question-example-2
User Sharifa
by
4.5k points
2 votes

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
User Abdul Raziq
by
4.3k points