Question

Solution of proce tgat tana-cota/sina*cosa=tan²a-coa

Answer 1

LHS = (tanA-cotA)/(sinAcosA)

= [(sinA/cosA)-(cosA/sinA)]/(sinAcosA)

= [sin²A-cos²A]/(sin²Acos²A)

= sin²A/(sin²Acos²A)-cos²A/(sin²Acos²A)

= 1/cos²A - 1/sin²A

= sec²A - cosec² A

= ( 1+tan²A )- ( 1 + cot²A )

= 1+tan²A-1-cot²A

= tan²A - cot²A

= RHS

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