Question

sin^2A/cos ^2A + cos^2A/sin^2A = 1/sin^2Acos^2A - 2​

Answer 1

Answer:

Step-by-step explanation:

LHS

sin2A/cos2A + cos2A/sin2A

by taking lcm,

sin4A + COS4a/sin2Acos2A

(sin2A)2 + (cos2A)2/sin2Acos2A

a2 + b2 = (a+b)2 - 2ab

(sin2A + cos2A)2 - 2sin2Acos2A/cos2Asin2A

1 - 2sin2Acos2A/cos2Asin2A

1/sin2Acos2A - 2cos2Asin2A/sin2Acos2A

1/sin2Acos2A- 2

hence proved

Answer 2

sin²A/cos²A + cos²A/sin²A = 1/(sin²A cos²A) - 2​

LHS:

sin²A/cos²A + cos²A/sin²A = (sin⁴A + cos⁴A)/(sin²A cos²A)

⇒ sin²A/cos²A + cos²A/sin²A = ((sin²A)² + (cos²A)²)/(sin²A cos²A)

∵ a² + b² = (a + b)² - 2ab

⇒ sin²A/cos²A + cos²A/sin²A = ((sin²A + cos²A)² - 2 sin²A cos²A)/(sin²A cos²A)

∵ sin²A + cos²A = 1

⇒ sin²A/cos²A + cos²A/sin²A = ((1)² - 2 sin²A cos²A)/(sin²A cos²A)

⇒ sin²A/cos²A + cos²A/sin²A = (1 - 2 sin²A cos²A)/(sin²A cos²A)

⇒ sin²A/cos²A + cos²A/sin²A = 1/(sin²A cos²A) - (2 sin²A cos²A)/(sin²A cos²A)

⇒ sin²A/cos²A + cos²A/sin²A = 1/(sin²A cos²A) - 2

∴ sin²A/cos²A + cos²A/sin²A = RHS

Hence proved.