Question

secA(1-sinA) (secA + tanA) =1. Prove that.​

Answer 1

Step-by-step explanation:

secA(1-sinA)(secA+tanA)

= (1/cosA)(1-sinA)(1/cosA+sinA/cosA)

= (1/cos²A)(1-sinA)(1+sinA)

= (1/cos²A)(1-sin²A)

= (1/cos²A)(cos²A)

= 1

= RHS

This is the correct proof.

Plzz mark as brainlist:-) :-) :-)

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