solve cos8A cos5A-cos12acos9A÷sin8A cos5A+cos12A sin9A=tan4A
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Step-by-step explanation:
(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
kalaisiva4132:
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