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Answer» Question is wrong because TANA × cot A = 1
the right question is (COSECA - sinA)(secA-cosA) = (1/(tanA+cotA)
SOLUTION: LHS = (cosecA-sinA)(secA-cosA) LHS = ((1/sinA)-sinA)((1/cosA)-cosA) LHS = ((1-sin²A)/sinA)((1-cos²A)/cosA) [NOTE: 1-sin²A = cos²A and 1-cos²A = sin²A] LHS = (cos²A/sinA)(sin²A/cosA) LHS = (cos²A sin²A)/(sinA cosA) LHS = cosA sinA/1 [note : 1 = sin²A+cos²A] LHS = cosA sinA/(sin²A+cos²A) LHS = 1/((sin²A/cosA sinA)+(cos²A/cosA sinA)) LHS = 1/((sinA/cosA)+(cosA/sinA) [note : sinA/cosA = tanA and cosA/sinA = cotA] LHS = 1/(tanA + cotA)
LHS=RHS Hence proved
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