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1+secA by secA=sin^2A by my 1-cosA

Answer» Solving LHS(1+secA)/secA × (1-cosA)(1/secA+1)(1-cosA)(CosA+1)(1-cosA)1-cos²A= sin²A = RHS
To prove 1+secA/secA = sin²A/1-cosA ,or(1+secA)(1-cosA)/secA=sin²A


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