Math, asked by Sourav007, 1 year ago

(sin8xcosx-sin6xcos3x)/(sin2xcosx-sin3xsin4x)

Answers

Answered by priya201709otqosa
31
1. 2 sinA cosB = sin (A+B) + sin (A-B) 
2. 2 cosA cosB = cos (A+B) + cos (A-B) 
3. 2 sinA sinB = cos (A-B) - cos (A+B) 
4. sinx / cosx = tanx 
5. sinC - sinD = 2 cos ( (C+D) / 2 ) sin ( (C-D) / 2 ) 
6. cosC + cosD = 2 cos ( (C+D) / 2 ) cos ( (C-D) / 2 ) 

(sin8xcosx - sin6xcos3x) / (cos2xcosx - sin3xsin4x) 
= 2 (sin8xcosx - sin6xcos3x) / 2 (cos2xcosx - sin3xsin4x) 
= (2sin8xcosx - 2sin6xcos3x) / (2cos2xcosx - 2sin3xsin4x) 
= (sin9x + sin7x - (sin 9x + sin3x)) / (cos3x + cosx - (cosx - cos7x)) 
= (sin7x - sin3x) / (cos7x + cos3x) 
= 2 cos5x sin2x / 2 cos5x cos2x 
= sin2x / cos2x 
= tan2x
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