Question

(sec2A- tan2A)= 1-sinA/ 1+SinA

Answer 1

Please see the attachment

Answer 2

RHS = 1-sinA/ 1+SinA

=[(1-sinA)/(1+sinA)][(1-sinA)/(1-sinA)]

= (1-sinA)²/[1²-sin²A]

= (1+sin²A-2sinA)/(cos²A)

= 1/cos²A+sin²A/cos²A-2sinA/cos²A

= sec²A+tan²A-2tanAsecA

= (secA-tanA)²

= LHS

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