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Grade: 12th pass
        
this is a question asked in mca entrance..i m nt able to tackle it in easy way...kindly help
6 months ago

Answers : (2)

Anish Singhal
askIITians Faculty
1200 Points
							Its quite simple just apply chain rule after simplifying the given expression(use basic formulas of trigonometry)
						
6 months ago
Rajat
180 Points
							
y'=2sinacosa-2sin(a+b)cos(a+b)+2sinb[cosacos(a+b)-sinasin(a+b)]
=sin2a-sin2(a+b)+2sinbcos(2a+b)
=sin[(2a+b)-b] - sin[(2a+b)+b] + 2sinbcos(2a+b)
= -2sinbcos(2a+b) + 2sinbcos(2a+b)
= 0 
So, y'''=0
 
Please approve and best of luck
6 months ago
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  • Discussion Forum
  • Previous Year Exam Questions


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