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CBSE Class 12 Math 2022 Term I Solved Paper

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Question : 4 of 50
Marks: +1, -0
If sin⁡y=xcos⁡(a+y)\sin y = x \cos(a+y), then dxdy\frac{dx}{dy} is
Solution:     👈: Video Solution
Explanation: Given, sin⁡y=xcos⁡(a+y)\sin y = x \cos(a+y)
⇒x=sin⁡ycos⁡(a+y)\Rightarrow \quad x = \frac{\sin y}{\cos(a+y)}
Differentiating with respect to yy, we get
dxdy=cos⁡(a+y)ddy(sin⁡y)−sin⁡yddy{cos⁡(a+y)}cos⁡2(a+y)\frac{dx}{dy} = \frac{ \cos(a+y) \frac{d}{dy}(\sin y) - \sin y \frac{d}{dy}\{\cos(a+y)\} }{ \cos^2(a+y) }
⇒dxdy=cos⁡(a+y)cos⁡y−sin⁡y[−sin⁡(a+y)]cos⁡2(a+y)\Rightarrow \frac{dx}{dy} = \frac{ \cos(a+y) \cos y - \sin y [-\sin(a+y)] }{ \cos^2(a+y) }
⇒dxdy=cos⁡(a+y)cos⁡y+sin⁡ysin⁡(a+y)cos⁡2(a+y)\Rightarrow \frac{dx}{dy} = \frac{ \cos(a+y) \cos y + \sin y \sin(a+y) }{ \cos^2(a+y) }
⇒dxdy=cos⁡[(a+y)−y]cos⁡2(a+y)\Rightarrow \frac{dx}{dy} = \frac{ \cos[(a+y)-y] }{ \cos^2(a+y) }
⇒dxdy=cos⁡2acos⁡2(a+y)\Rightarrow \frac{dx}{dy} = \frac{ \cos^2 a }{ \cos^2(a+y) }
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