CBSE Class 12 Math 2022 Term I Solved Paper

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Question : 4 of 50
 
Marks: +1, -0
If siny=xcos(a+y), then dxdy is
Solution:
Explanation: Given, siny=xcos(a+y)
⇒x=sinycos(a+y)
Differentiating with respect to y, we get
dxdy=cos(a+y)ddy(siny)−sinyddy{cos(a+y)}cos2(a+y)
⇒dxdy=cos(a+y)cosy−siny[−sin(a+y)]cos2(a+y)
⇒dxdy=cos(a+y)cosy+sinysin(a+y)cos2(a+y)
⇒dxdy=cos[(a+y)−y]cos2(a+y)
⇒dxdy=cos2acos2(a+y)
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