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Cos(a-b)=5/13 & sin(a+b)=4/5 then find the value Sin2b...plz make with explaination ...................

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2 years ago

Navneet kumar
13 Points
```							a-b = arccos (5/13) ............(1)a+b = arcsin (4/5)  ..............(2)(2) - (1)2b = arcsin(4/5) - arccos (5/13)2b = pi/2 - arccos(4/5) - arccos(5/13)Arccos (4/5) + arccos (5/13) = pi/2 - 2barccos (4/5×5/13 - sqroot[1 - (4/5)^2] x sqroot[1 - (5/13)^2] = pi/2 - 2b20/65 - sqroot ( 1- 16/25)×sqroot(1-25/169) = cos(pi/2-2b)20/65 - sqroot(9/25)×sqroot(144/169) = sin(2b)20/65 - 3/5×12/13 = sin(2b)(20-36)/65 = sin(2b)》sin2b = -16/65
```
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