Question

Prove that sec A (1 - sin A) (sec A + tan A) = 1​

Answer 1

Step-by-step explanation:

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