prove that (sin A +cosec A)2 +(cosA + secA )= 7 + tan2A + cot2A
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By using the algebra identity, (a + b)² = a² + b² + 2ab,
Left side is: sin²A + csc²A + 2 + cos²A + sec²A + 2
[Since sinA*cscA = 1 = cosA*secA]
= sec²A + csc²A + 5 [Since sin²A + cos²A = 1]
= tan²A + cot²A + 7 [Since sec²A = 1 + tan²A and csc²A = 1 + cot²A]
Thus one possible result is:
(sinA + cscA)² + (cosA + secA)² = tan²A + cot²A + 7
Left side is: sin²A + csc²A + 2 + cos²A + sec²A + 2
[Since sinA*cscA = 1 = cosA*secA]
= sec²A + csc²A + 5 [Since sin²A + cos²A = 1]
= tan²A + cot²A + 7 [Since sec²A = 1 + tan²A and csc²A = 1 + cot²A]
Thus one possible result is:
(sinA + cscA)² + (cosA + secA)² = tan²A + cot²A + 7
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