If x cos theta + y sin theta = 5 and x sin th | vedclass

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(D) Given equations are:$1) x \cos 	heta + y \sin 	heta = 5$$2) x \sin 	heta - y \cos 	heta = 3$Squaring both equations:$(x \cos 	heta + y \sin \

Vedclass · Accepted Answer

$34$

Vedclass · Answer

(D) Given equations are:$1) x \cos 	heta + y \sin 	heta = 5$$2) x \sin 	heta - y \cos 	heta = 3$Squaring both equations:$(x \cos 	heta + y \sin 	heta)^{2} = 5^{2} \implies x^{2} \cos^{2} 	heta + y^{2} \sin^{2} 	heta + 2xy \sin 	heta \cos 	heta = 25$$(x \sin 	heta - y \cos 	heta)^{2} = 3^{2} \implies x^{2} \sin^{2} 	heta + y^{2} \cos^{2} 	heta - 2xy \sin 	heta \cos 	heta = 9$Adding the two squared equations:$(x^{2} \cos^{2} 	heta + x^{2} \sin^{2} 	heta) + (y^{2} \sin^{2} 	heta + y^{2} \cos^{2} 	heta) = 25 + 9$$x^{2}(\cos^{2} 	heta + \sin^{2} 	heta) + y^{2}(\sin^{2} 	heta + \cos^{2} 	heta) = 34$Since $\sin^{2} 	heta + \cos^{2} 	heta = 1$,we get:$x^{2}(1) + y^{2}(1) = 34$$x^{2} + y^{2} = 34$

List-$A$	List-$B$
$(I)$ $\tan \left(\frac{\alpha + \beta}{2}\right) =$	$(a)$ $\frac{b}{a}$
$(II)$ $\cos (\alpha + \beta) =$	$(b)$ $\frac{2ab}{a^2 + b^2}$
$(III)$ $\sin (\alpha + \beta) =$	$(c)$ $\frac{2ab}{a^2 - b^2}$
$(IV)$ $\tan (\alpha + \beta) =$	$(d)$ $\frac{a^2 - b^2}{a^2 + b^2}$

If $x \cos \theta + y \sin \theta = 5$ and $x \sin \theta - y \cos \theta = 3$,then the value of $x^{2} + y^{2}$ is:

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