(tan57° + cot37°)/ (tan33° + cot53° ) =?

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(tan57° + cot37°)/ (tan33° + cot53° ) =?

Asin53° + cos33°

Btan53°× tan57°

Csin53° × sin57°

Dcos57° × cos53°

Answer:

B. tan53°× tan57°

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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°

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