Question

Solve That

1+sec A/Sec A = sin^2 A/1-cos A​

Answer 1

$\huge\boxed{Answer}$

(1 + SecA)/SecA

(1/SecA) + (SecA/SecA)

CosA + 1

1 + CosA

From rationalisation

(1 + CosA) × (1 - CosA)/(1 - CosA)

(1 - Cos²A)/(1 - CosA)

Sin²A/(1 - CosA)

Answer 2

$\frac{1 + \sec(a) }{ \sec(a) }$

$= > \frac{1}{ \sec(a) } + \frac{ \sec(a) }{ \sec(a) }$

$= > \cos(a) + 1$

multiplying by 1-cos(A)/1-cos(A)

$\frac{(1 + \cos(a) )(1 - \cos(a)) }{1 - \cos(a) }$

$\frac{1 - {cos}^{2} (a)}{1 - \cos(a) }$

$\frac{ {sin}^{2} (a)}{1 - \cos(a) }$