If secA = 17/8 prove that (3-4sin 2 A) / (4cos 2 A-3) = (3-tan 2 A) / (1-3tan 2 A)
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Sec A = 17/8
(3 - 4 Sin²A) / (4 Cos²A - 3)
= [1 + 2 (1 - 2 sin²A)] / [2(2 cos²A - 1) - 1]
= [ 1 + 2 cos2A ] / [ 2 Cos2A - 1]
= [ 1 + 2 (1 - tan²A)/(1+tan²A) ] / [ 2(1-tan²A)/(1+ tan²A) - 1]
= [ 3 - tan²A ] / [ 1 - 3 tan²A ]
(3 - 4 Sin²A) / (4 Cos²A - 3)
= [1 + 2 (1 - 2 sin²A)] / [2(2 cos²A - 1) - 1]
= [ 1 + 2 cos2A ] / [ 2 Cos2A - 1]
= [ 1 + 2 (1 - tan²A)/(1+tan²A) ] / [ 2(1-tan²A)/(1+ tan²A) - 1]
= [ 3 - tan²A ] / [ 1 - 3 tan²A ]
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