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Question : 67 of 160
Marks:
+1,
-0
Solution:
It is given that,
a≠b,x≠nπ,n∈z and
y2=a2cos2x+b2sin2x Differentiate the above equation with respect to
x 2y=−a2sin2x+b2sin2x 2y=(b2−a2)sin2x Again differentiate the above equation with respect to
x 2(y+()2)=2(b2−a2)cos2x y+()2=(b2−a2)cos2x Multiply
y2 on both sides,
y3+y2()2=y2(b2−a2)cos2x y3=y2(b2−a2)cos2x−y2()2 Add
y4 in the above equation.
y4+y3=y4+y2(b2−a2)cos2x−y2()2+y4 Substitute the values in the above equation.
y4+y3=y4+y2(b2−a2)cos2x−y2()2+y4 =(a2cos2x+b2sin2x)(b2−a2)(cos2x−sin2x)−(b2−a2sinxcosx)2+(a2cos2x+b2sin2x)2 =a2b2cos4x+a2b2sin2x+2a2b2sin2xcos2x =a2b2(sin2x+cos2x)2 =a2b2 Or it can be written as
+y=()2
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