1 + cotA/secA = sin²A /1 - cosA
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Answered by
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Answer:
(1 + secA)/SecA = sin²A/(1-CosA)
Step-by-step explanation:
1 + cotA/secA = sin²A /1 - cosA
Correct Question is
(1 + secA)/SecA = sin²A/(1-CosA)
LHS
= (1+SecA)/SecA
SecA = 1/CosA
= (1 + 1/CosA)/(1/CosA)
= CosA + 1
RHS
= Sin²A/(1-CosA)
Sin²A = 1 - Cos²A
= ( 1 - Cos²A)/(1-CosA)
= (1 + CosA)(1-CosA) /(1-CosA)
= 1 + CosA
LHS = RHS
or question can be
1 + cotA/CosecA = Sin²A/(1-CosA)
LHS
= 1 + (CosA/SinA)/(1/SinA) ( cotA = cosA/SinA & CosecA = 1/SinA)
= 1 + CosA
RHS
= Sin²A/(1-CosA)
Sin²A = 1 - Cos²A
= ( 1 - Cos²A)/(1-CosA)
= (1 + CosA)(1-CosA) /(1-CosA)
= 1 + CosA
LHS = RHS
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Answer:
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