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Byju's Answer
Standard XII
Mathematics
Basic Inverse Trigonometric Functions
Cos8Acos5A-co...
Question
Cos8Acos5A-cos12Acos9A
____________________________
Sin8Acos5A+cos12Asin9A
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Solution
(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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