Question

Prove that :

Answer 1

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:

Answer 2

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