Question

prove that:

secA(1 - sinA) (secA + tanA) = 1​

Answer 1

Answer:

secA (1-sinA) (secA + tan A)  

= (secA-sinA×secA)(secA+tanA)

=(secA-tanA) (secA+tanA)  as secA=1/cosA and sinA/cosA=tanA

= (sec²A-tan²A) as (a+b)(a-b)=a²-b²

=sec²A-tan²A=1 from identity.

Hence proved.

Explanation:

Answer 2

LHS

$secA(1-sinA)(secA+tanA) \\ \\ = \frac{1}{cosA} (1-sinA)(secA+tanA) \\ \\ = ( \frac{1}{cosA} - \frac{ sinA}{cosA} )(secA+tanA) \\ \\ = (secA-tanA)(secA+tanA) \\ \\ = {sec}^{2} A - {tan}^{2} A\\ \\ = 1$

RHS

=1

LHS=RHS