Question

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

Answer 1

secA(1 - sinA) (secA + tanA)
(secA - secA.sinA) (secA + tanA)
(secA - sinA/cosA) (secA +tanA)
(secA - tanA) (secA + tanA)
(sec^2A - tan^2A)
because { sec^2A = 1 + tan^2A }
then (sec^2A - tan^2A) = 1.

Answer 2

Step-by-step explanation:

$\huge{\pink{\boxed{\green{\underline{\red{\sf{SOLUTION-}}}}}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {\blue{to \: prove}} \\ { \purple{ \boxed{ \red{ secA (1 - sinA)( secA + tanA) = 1 }}}}$

☆ According to given question:

☆ PROOF:

$\to secA (1 - sinA)( secA + tanA) = 1 \\ lhs \\ \to secA (1 - sinA )( secA + tanA ) \\ \to (\frac{1}{ cosA}) (1 - sinA )( frac{1}{ cosA } + \frac{ sinA }{ cosA } ) \\ \to \frac{(1 - sinA)(1 + sinA) }{ cos^{2} A } \\ \to \frac{1 - sin ^{2} A }{ cos^{2} A } \\ \to \frac{ cos ^{2} A }{ cos ^{2} A } \\ \to 1$

$\huge{\pink{\boxed{\green{\underline{\sf{PROVED:}}}}}}$