Question

prove that sec A [1-sin A] [sec A+tan A]=1​

Answer 1

sec A = 1/cos A

sec A (1-sin A) = secA - sinA/cosA

= sec A - tanA

= (1- SinA)/ cosA -----P

secA+tan A = (1+sinA)/ cosA -----Q

P×Q =

(1-sinA)(1+sinA)/cos²A

= 1-sin²A /cos²A

= cos²A/cos²A =1

Answer 2

Step-by-step explanation:

RHS = secA ( 1- sinA ) (secA + tanA )

{ since secA = 1/cosA, tanA = sinA/ cosA }

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

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

= secA (1- sin^2 A) / cosA

= (secA * cos^2 A)/ cos A

= cos^2 A / cos^2 A

= 1

RHS = LHS

Hence proved.