If x=a(sin theta+cos theta) and y=b(sin theta | vedclass

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(C) Given: $x=a(\sin 	heta+\cos 	heta)$ and $y=b(\sin 	heta-\cos 	heta)$.Dividing by $a$ and $b$ respectively,we get $\frac{x}{a}=\sin 	heta+\cos

Vedclass · Accepted Answer

$2$

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(C) Given: $x=a(\sin 	heta+\cos 	heta)$ and $y=b(\sin 	heta-\cos 	heta)$.Dividing by $a$ and $b$ respectively,we get $\frac{x}{a}=\sin 	heta+\cos 	heta$ and $\frac{y}{b}=\sin 	heta-\cos 	heta$.Now,consider the expression $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$.Substituting the values,we get $(\sin 	heta+\cos 	heta)^{2}+(\sin 	heta-\cos 	heta)^{2}$.Expanding the squares using $(A+B)^2 = A^2+B^2+2AB$ and $(A-B)^2 = A^2+B^2-2AB$,we get:$(\sin^2 	heta + \cos^2 	heta + 2\sin 	heta \cos 	heta) + (\sin^2 	heta + \cos^2 	heta - 2\sin 	heta \cos 	heta)$.Since $\sin^2 	heta + \cos^2 	heta = 1$,the expression simplifies to $1 + 2\sin 	heta \cos 	heta + 1 - 2\sin 	heta \cos 	heta$.Thus,the result is $1 + 1 = 2$.

If $x=a(\sin \theta+\cos \theta)$ and $y=b(\sin \theta-\cos \theta)$,then the value of $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$ is

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