Math, asked by payaljha91725, 5 months ago

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

Answers

Answered by Anonymous

6

sin8Acos5A+cos12Asin9A

cos8Acos5A−cos12Acos9A

=

2

1

[(sin13A+sin3A)+(sin21Asin3A)]

2

1

[(cos13A+cos3A)−(cos21Acos3A)]

=

sin13A+sin21A

cos13A−cos21A

=

2sin(

2

13+21

)Acos(

2

21−13

)A

2sin(

2

13+21

)Asin(

2

21−13

)A

=

cos4A

sin4A

=tan4A

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