Question

Prove that (sec theta - tan theta)square = 1- sin theta /1+sin theta

Answer 1

Answer:

Step-by-step explanation:

Answer 2

To prove ------>

( Secθ - tanθ )² = ( 1 - Sinθ ) / ( 1 + Sinθ )

Proof -----> LHS

= ( Secθ - tanθ )²

We know that,

Secθ = 1 / Cosθ , tanθ = Sinθ / Cosθ , applying these formulee here, we get,

= ( 1 / Cosθ - Sinθ / Cosθ )²

= { ( 1 - Sinθ ) / Cosθ }²

= ( 1 - Sinθ )² / Cos²θ

We know that , Cos²A = 1 - Sin²A , applying it , we get,

= ( 1 - Sinθ )² / ( 1 - Sin²θ )

= ( 1 - Sinθ )² / ( 1 )² - ( Sinθ )²

We know that, a² - b² = ( a + b ) ( a - b ) , applying it , we get,

= ( 1 - Sinθ )² / ( 1 + Sinθ ) ( 1 - Sinθ )

= ( 1 - Sinθ ) / ( 1 + Sinθ ) = RHS