Anyone pls answer the 28 and 29th URGENTLY I will mark you as the Brainly answer
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Answer:
Given xSinA – yCosA = 0
⇒ x.SinA = y CosA ........(1)
Consider xSin^3A+yCos^3A = SinACosAxSin
3
A+yCos
3
A=SinACosA
xSin^3A+yCosACos^2A = SinACosAxSin
3
A+yCosACos
2
A=SinACosA
xSin^3A+ xSinACos^2A = SinACosAxSin
3
A+xSinACos
2
A=SinACosA {From eq 1}
xSin^3A+ xSinA(1-Sin^2A) = SinACosAxSin
3
A+xSinA(1−Sin
2
A)=SinACosA
xSin^3A+ xSinA-xSin^3A = SinACosAxSin
3
A+xSinA−xSin
3
A=SinACosA
xSinA = SinACosA
x = CosA ......(2)
Substitute this value in the eq (1)
We get y = SinA
so, x^2 + y^2 = Cos^2A + Sin^2A = 1x
2
+y
2
=Cos
2
A+Sin
2
A=1
PLS MARK ME AS BRAINLIEST
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