If ( r - 3 )^2 + ( r - 1 )^2 = 0, then the value of r + r + is equal to?

Answer: Option A

$$\begin{array}{rl}& {\left(r\cos \theta -\sqrt{3}\right)}^2+{\left(r\sin \theta -1\right)}^2=0\\ & \Rightarrow {\left(r\cos \theta -\sqrt{3}\right)}^2=0,{\left(r\sin \theta -1\right)}^2=0\\ & \Rightarrow r\text{ cos}\theta =\sqrt{3}........(i)\\ & \Rightarrow r\sin \theta =\mathrm{1.........}(ii)\\ & \text{Squaring and adding equation (i) and (ii)}\\ & \Rightarrow r^2{\cos }^2\theta +r^2{\sin }^2\theta =3+1\\ & \Rightarrow r^2\left(\text{co}{\text{s}}^2\theta +{\sin }^2\theta \right)=4\\ & \Rightarrow r^2=4\\ & \Rightarrow r=2\\ & \text{tan}\theta =\frac{r\sin \theta }{r\text{ cos}\theta }=\frac{1}{\sqrt{3}}\text{ and }r\text{ cos}\theta =\sqrt{3}\\ & \cos \theta =\frac{\sqrt{3}}{r},\\ & \sec \theta =\frac{r}{\sqrt{3}}\\ & \frac{r\tan \theta +\sec \theta }{r\sec \theta +\tan \theta }\\ & =\frac{\frac{r}{\sqrt{3}}+\frac{r}{\sqrt{3}}}{\frac{r^2}{\sqrt{3}}+\frac{1}{\sqrt{3}}}\\ & =\frac{r\left(\frac{2}{\sqrt{3}}\right)}{\frac{r^2+1}{\sqrt{3}}}\\ & =\frac{2r}{r^2+1}\\ & =\frac{2\times 2}{2^2+1}\\ & =\frac{4}{5}\\ & \\ & \mathbf{Alternate}:\\ & r=2\\ & \text{tan}\theta =\frac{r\sin \theta }{r\text{ cos}\theta }=\frac{1}{\sqrt{3}}\\ & \theta =30^{\circ }\\ & =\frac{2\tan {30}^{\circ }+\sec {30}^{\circ }}{2\sec {30}^{\circ }+\tan {30}^{\circ }}\\ & =\frac{2\times \frac{1}{\sqrt{3}}+\frac{2}{\sqrt{3}}}{2\times \frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}}}\\ & =\frac{4}{5}\end{array}$$