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```
cos8A cos5A - cos12A cos9A / sin8A cos5A + cos12A sin9A = tan4A
cos8A cos5A - cos12A cos9A / sin8A cos5A + cos12A sin9A = tan4A

```
2 years ago

Vivek kumar
15 Points
```							Because tanA=sinA/cosABy using identies of the chapter 9 some application of trigonometry you will be able to find out than the equation Cos8A cos5A - cos12A cos9A/sin8A cos5A + cos12A sin9A = tan4A
```
2 years ago
Arun
25768 Points
```							(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  RegardsArun (askIITians forum expert)
```
2 years 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