Question

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

Answer 1

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