Question

(CosecA+SinA)(SecA-CosA)=1/tanAcotA
please solve this one for me

Answer 1

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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