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this is a question asked in mca entrance..i m nt able to tackle it in easy way...kindly help

this is a question asked in mca entrance..i m nt able to tackle it in easy way...kindly help

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Grade:12th pass

2 Answers

Anish Singhal
askIITians Faculty 1192 Points
5 years ago
Its quite simple just apply chain rule after simplifying the given expression(use basic formulas of trigonometry)
Rajat
213 Points
5 years ago
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

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