Question

(secA+tanA)(1-sinA)=?​

Answer 1

Answer:

cos²A

Step-by-step explanation:

$( \sec \alpha + \tan \alpha )(1 - \sin \alpha ) \\ = \sec \alpha - \sec \alpha \sin \alpha + \tan \alpha - \tan \alpha \sin \alpha \\ = \frac{1}{ \cos \alpha } - ( \frac{1}{ \cos \alpha } \times \sin \alpha ) + \frac{ \sin \alpha }{ \cos \alpha } - ( \frac{ \sin \alpha }{ \cos \alpha } \times \sin \alpha ) \\ = \frac{1}{ \cos \alpha } - \frac{ { \sin }^{2} \alpha }{ \cos \alpha } \\ = \frac{(1 - { \sin }^{2} \alpha ) }{ \cos \alpha } \\ = \frac{ { \cos }^{2} \alpha }{ \cos \alpha } \\ = \cos \alpha$

IDENTITY USED:

sin²x + cos²x = 1

HOPE IT HELPS :D