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`        14 and 15 no plsMultiple angle.                                          .........................................................`
11 months ago

## Answers : (1)

1673 Points
```							 32 sin^6x = 4.(2sin^2 x)^3 = 4(1- cos2x)^3 = 4(1-cos^3 2x -3cos 2x +3cos^2 2x) = 4- 12cos 2x + 6(1+cos 4x) – (cos 6x+ 3cos2x) = 10 – 15cos 2x +6cos 4x -cos6xthen,10- 15cos 2x +6cos 4x – cos 6x =10 – 15cos2x + b cos4x+a cos6xor (6-b)cos4x = (1+a)cos6xcos4x/cos6x = (1+a)/(6-b)componendo ad dividendoor (cos4x – cos 6x)/(cos 4x + cos6x) = (1+a-6+b)/(1+a+6-b)or tan5x tanx =(a+b-5)/(7+a-b)
```
11 months ago
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### Course Features

• 31 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions