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) Quest…

Question

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`

Answer 1

Please find attached herewith the solutions of your questions.

Hope it helps.

Answer 2

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