Question

. sec A (1 - sin A) (sec A + tan A) = 1​

Answer 1

Question:-

$\sec A(1 - \sin A )( \sec A + \tan \: A) = 1$

Solution:-

Using this trigonometry identities

$\rm \: \to \sec(A) = \frac{1}{ \cos A }$

$\to \rm \: \tan(A) = \frac{ \sin(A) }{ \cos(A) }$

We get

$\rm \: \frac{1}{ \cos A } (1 - \sin \: A)( \frac{1}{ \cos A } + \frac{ \sin A }{ \cos A } )$

$\rm \: \frac{1 - \sin(A) }{ \cos(A) } \times \frac{1 + \sin(A) }{ \cos(A) }$

Using this identity

$\to \: \rm \: (a - b)(a + b) = ( {a}^{2} - {b}^{2} )$

$\rm \: \frac{1 - \sin {}^{2} (A) }{ \cos {}^{2} (A) }$

Trigonometry identity

$\rm \: 1 - \sin {}^{2} (A) = \cos {}^{2} (A)$

$\rm \: \frac{ \cos {}^{2} (A) }{ \cos {}^{2} (A) }$

$= 1$

Hence proved