Math, asked by devkaushik71, 1 year ago

cos5A.cos8A-cos12A.cos9A/sin8A.cos5A+cos12A.sin9A

Answers

Answered by brainlyuttkarsh

(cos8A.cos5A-cos12A.cos9A)/(sin8A.cos5A+cos12A.sin9A)

=> [1/2 {cos13A+cos3A}-1/2 {cos21A+cos3A}]/[1/2 {sin13A+sin3A}+1/2 {sin21A-sin3A}]

=>( cos13A-cos21A)/(sin13A+sin21A)

=> 2sin17A.sin4A/2sin17A.cos4A

=> tan4A

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