$\displaystyle \mathrm{P}(x, y), \mathrm{Q}(-2,-3)$ and $\displaystyle \mathrm{R}(2,3)$ are the vertices of a right triangle PQR right angled at P. Find the relationship between $\displaystyle x$ and $\displaystyle y$. Hence, find all possible values of $\displaystyle x$ for which $\displaystyle y=2$.

In \(\displaystyle \Delta \mathrm{PQR}, \angle \mathrm{P}=90^{\circ}\) \[\begin{aligned} & P Q^{2}+P R^{2}=Q R^{2} \\ & \Rightarrow(x+2)^{2}+(y+3)^{2}+(x-2)^{2}+(y-3)^{2}=4^{2}+6^{2} \\ & \Rightarrow x^{2}+4 x+4+y^{2}+6 y+9+x^{2}-4 x+4+y^{2}-6 y+9=52 \end{aligned} \] gives, \(\displaystyle x^{2}+y^{2}=13\) Now for \(\displaystyle y=2, x= \pm 3\) \[\begin{aligned} \mathrm{LHS} & =\frac{\cos \mathrm{A}+\sin \mathrm{A}-1}{\cos \mathrm{~A}-\sin \mathrm{A}+1} \\ & =\frac{\cot \mathrm{A}+1-\operatorname{cosec} \mathrm{A}}{\cot \mathrm{~A}-1+\operatorname{cosec} \mathrm{A}} \\ & =\frac{\cot \mathrm{A}-\operatorname{cosec} \mathrm{A}+\operatorname{cosec}^{2} \mathrm{~A}-\cot ^{2} \mathrm{~A}}{\cot \mathrm{~A}-1+\operatorname{cosec} \mathrm{A}} \\ & =\frac{(\operatorname{cosec} \mathrm{A}-\cot \mathrm{A})(-1+\operatorname{cosec} \mathrm{A}+\cot \mathrm{A})}{\cot \mathrm{A}-1+\operatorname{cosec} \mathrm{A}} \\ & =\operatorname{cosec} \mathrm{A}-\cot \mathrm{A}=\mathrm{RHS} \end{aligned} \]\[\begin{aligned} & \text { LHS }=p^{2}-q^{2} \\ & =(\cot \theta+\cos \theta)^{2}-(\cot \theta-\cos \theta)^{2} \\ & =[(\cot \theta+\cos \theta)+(\cot \theta-\cos \theta)][(\cot \theta+\cos \theta)-(\cot \theta-\cos \theta)] \\ & =2 \cot \theta \times 2 \cos \theta=4 \cot \theta \cos \theta \\ & \text { RHS }=4 \sqrt{p q} \\ & =4 \sqrt{(\cot \theta+\cos \theta)(\cot \theta-\cos \theta)} \\ & =4 \sqrt{\cot ^{2} \theta-\cos ^{2} \theta} \\ & =4 \sqrt{\cos ^{2} \theta\left(\operatorname{cosec}^{2} \theta-1\right)} \\ & =4 \sqrt{\cos ^{2} \theta \times \cot ^{2} \theta} \\ & =4 \cot \theta \cos \theta \\ & \therefore \text { LHS }=\text { RHS } \end{aligned} \]