Question

sec A + tan A- 1 / tanA - secA+ 1 = sec A ( 1+ sin A )​

Answer 1

(secA + tanA - 1)/(tanA - secA + 1) = sec(1 + sinA)

LHS

(secA + tanA - 1)/(tanA - secA + 1)

substitute that 1 in the numerator by

sec^2A = 1 + tan^2A

sec^2A - tan^2A = 1

{(secA + tanA - (sec^2A - tan^2A)}/(tanA - secA + 1)

{(secA + tanA) - (secA + tanA).(secA - tanA)}/(tanA - secA + 1)

Now, take (secA + tanA) common

(secA + tanA){1 - (secA - tanA)} / (tanA - secA + 1)

(secA + tanA){1 - secA + tanA} / (tanA - secA + 1)

secA + tanA

1/cosA + sinA/cosA

(1 + sinA) / cosA

secA(1 + sinA)