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

Question

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

Aicharya · Accepted Answer

$\begin{aligned} & \text {Squaring both given equations, } \ & x^2 \cos ^2 \theta+y^2 \sin ^2 \theta+2 x y \cos \theta \sin \theta=25 \ & \Rightarrow x^2 \cos ^2 \theta+y^2 \sin ^2 \theta+x y \sin 2 \theta=25 \ldots . \text { (i) } \ & x^2 \sin ^2 \theta+y^2 \cos ^2 \theta-2 x y \cos \theta \sin \theta=9 \ & \Rightarrow x^2 \sin ^2 \theta+y^2 \cos ^2 \theta-x y \sin 2 \theta=9 \ldots \text { (ii) } \end{aligned}$ $\begin{aligned} & \text {(i) }+(\mathrm{ii}), \ & x^2+y^2=34\end{aligned}$

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

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