Sin 2 A+cos 2 B+sec 2 C=tan 2 D given:sinA=cosB
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Given sin A = cos B = sin(π/2 - B)
=> A + B = π/2
Since A, B, C and D are angles of a quadrilateral. So
angles C+ D = 3π/2
=> cos (C+D) = 0
=> cos C cos D = sin C sin D
=> tan C = Cot D (1)
LHS = sin² A + cos² B + sec² C
= 2 sin² A + tan² C + 1
= 2 sin² A + Cot² D + 1 using (1)
This expression can be simplified only if there is a relation between the angles A and D.
=> A + B = π/2
Since A, B, C and D are angles of a quadrilateral. So
angles C+ D = 3π/2
=> cos (C+D) = 0
=> cos C cos D = sin C sin D
=> tan C = Cot D (1)
LHS = sin² A + cos² B + sec² C
= 2 sin² A + tan² C + 1
= 2 sin² A + Cot² D + 1 using (1)
This expression can be simplified only if there is a relation between the angles A and D.
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