Question

if sec A - tan A =3, then find the value of sec A + tan A

Answer 1

We know the identity
${ \sec }^{2} A - { \tan }^{2} A = 1 \\ ( \sec A + \tan A)( \sec A - \tan A) = 1 \\ \sec A + \tan A = \frac{1}{ \sec A - \tan A} = \frac{1}{3}$

Answer 2

We know sec²A - tan²A = 1




Now,


= > sec²A - tan²A = 1




By using a² - b² = ( a + b ) ( a - b )




= > ( secA+ tanA ) ( secA - tanA ) = 1


= > ( secA + tanA ) ( 3 ) = 1


= > ( secA + tanA ) × 3 = 1


$= > \text{secA + tanA } = \frac{1}{3}$