If x=frac{sin^3 theta}{cos^2 theta} and y=fra | vedclass

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(D) Given: $x=\frac{\sin^3 	heta}{\cos^2 	heta}$ and $y=\frac{\cos^3 	heta}{\sin^2 	heta}$.We need to find $x+y = \frac{\sin^3 	heta}{\cos^2 	he

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$\frac{79}{18}$

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(D) Given: $x=\frac{\sin^3 	heta}{\cos^2 	heta}$ and $y=\frac{\cos^3 	heta}{\sin^2 	heta}$.We need to find $x+y = \frac{\sin^3 	heta}{\cos^2 	heta} + \frac{\cos^3 	heta}{\sin^2 	heta} = \frac{\sin^5 	heta + \cos^5 	heta}{\sin^2 	heta \cos^2 	heta}$.Alternatively,$x+y = \frac{\sin^3 	heta \sin^2 	heta + \cos^3 	heta \cos^2 	heta}{\sin^2 	heta \cos^2 	heta} = \frac{\sin^5 	heta + \cos^5 	heta}{(\sin 	heta \cos 	heta)^2}$.Given $\sin 	heta + \cos 	heta = \frac{1}{2}$.Squaring both sides: $(\sin 	heta + \cos 	heta)^2 = (\frac{1}{2})^2$ $\Rightarrow 1 + 2 \sin 	heta \cos 	heta = \frac{1}{4}$ $\Rightarrow \sin 	heta \cos 	heta = -\frac{3}{8}$.Now,$x+y = \frac{\sin^3 	heta}{\cos^2 	heta} + \frac{\cos^3 	heta}{\sin^2 	heta} = \frac{\sin^5 	heta + \cos^5 	heta}{\sin^2 	heta \cos^2 	heta}$.Using $\sin^5 	heta + \cos^5 	heta = (\sin 	heta + \cos 	heta)(\sin^4 	heta - \sin^3 	heta \cos 	heta + \sin^2 	heta \cos^2 	heta - \sin 	heta \cos^3 	heta + \cos^4 	heta)$.$= (\sin 	heta + \cos 	heta)((\sin^2 	heta + \cos^2 	heta)^2 - 2 \sin^2 	heta \cos^2 	heta - \sin 	heta \cos 	heta(\sin^2 	heta + \cos^2 	heta) + \sin^2 	heta \cos^2 	heta)$.$= (\frac{1}{2})(1 - \sin^2 	heta \cos^2 	heta - \sin 	heta \cos 	heta)$.Substitute $\sin 	heta \cos 	heta = -\frac{3}{8}$:$x+y = \frac{\frac{1}{2}(1 - (-\frac{3}{8})^2 - (-\frac{3}{8}))}{(-\frac{3}{8})^2} = \frac{\frac{1}{2}(1 - \frac{9}{64} + \frac{3}{8})}{\frac{9}{64}} = \frac{\frac{1}{2}(\frac{64-9+24}{64})}{\frac{9}{64}} = \frac{\frac{1}{2} 	imes 79}{9} = \frac{79}{18}$.

If $x=\frac{\sin^3 \theta}{\cos^2 \theta}$ and $y=\frac{\cos^3 \theta}{\sin^2 \theta}$,where $\sin \theta+\cos \theta=\frac{1}{2}$,then $x+y=$

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