(tan57° + cot37°)/ (tan33° + cot53° ) =?Asin53° + cos33°Btan53°× tan57Csin53° × sin57°Dcos57° × cos53°Answer: B. tan53°× tan57 Read Explanation: Solution: Given: (tan57° + cot37°)/ (tan33° + cot53°) We know cot(A) = tan(90-A) So, [tan 33° = cot 57° and cot37° = tan 53° ] ⇒ (tan57° + tan 53°) / (cot 57° + cot 53°) ⇒ (tan57° + tan 53°) / (1/tan 57° + 1/tan 53°) ⇒ (tan57° + tan53°) / {(tan53° + tan57°)/tan53°× tan57°} ⇒ tan53° × tan57°Read more in App