Question

please help !


If sec theta + tan theta = k then prove that ( k^2 + 1 ) sin theta = k ^2 - 1 ​

Answer 1

=> sec²A - tan²A = 1 (identity)

=> (secA + tanA)(secA - tanA) = 1

=> (k)(secA - tanA) = 1

=> (secA - tanA) = 1/k

Adding secA + tanA and secA - tanA:

=> 2secA = k + 1/k

=> 2/cosA = (k² + 1)/k

=> 2k/(k² + 1) = cosA

We know, sin²A = 1 - cos²A. Thus,

=> sin²A = 1 - [ 2k/(k² + 1) ]²

=> sin²A = [(k² + 1)² - (2k)²]/(k² + 1)²

=> (k² + 1)²sin²A = (k⁴ + 1 + 2k² - 4k²)

=> (k² + 1)²sin²A = (k² - 1)²

=> (k² + 1)sinA = k² - 1

Answer 2

Step-by-step explanation:

please write identities used in this yourself