Question

1+sinA/1-sinA=(secA+tanA)^2​

Answer 1

Step-by-step explanation:

√1-sinA/1+sinA=√(1+sinA)²/cos²A

√1-sinA/1+sinA=1+sinA/cosA

√1-sinA/1+sinA/secA+tanA

1-sinA/1+sinA=(secA+tanA)²

Answer 2

Step-by-step explanation:

identities used:

a. 1/cosA = secA

b. sinA/cosA = tanA

c. (a-b)(a+b) = a2-b2

d. 1+tan2A=sec2A

therefore, sec2A-tan2A=1