Question : If $(r\cos \theta -\sqrt{3})^{2}+(r\sin \theta -1)^{2}=0$, then the value of $\frac{r\tan \theta +\sec \theta}{r\sec \theta +\tan\theta}$ is equal to:
Option 1: $\frac{4}{5}$
Option 2:
$\frac{5}{4}$
Option 3:
$\frac{\sqrt{3}}{4}$
Option 4:
$\frac{\sqrt{5}}{4}$
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Correct Answer: $\frac{4}{5}$
Solution : $(r\cos \theta -\sqrt{3})^{2}+(r\sin \theta -1)^{2}=0$ $⇒(r\cos \theta -\sqrt{3})^{2}=0$ and $(r\sin \theta -1)^{2}=0$ $⇒r\cos\theta =\sqrt3$ and $r\sin \theta =1$ $\therefore\tan\theta = \frac{1}{\sqrt3}$ Now, $r^2\sin^2\theta + r^2\cos^2\theta =4$ $⇒r^2 (\sin^2\theta + \cos^2\theta) =4$ Using identity $\sin^2\theta + \cos^2\theta=1$, we get, $⇒r^2 =4$ $⇒r= 2$ Using identity: $1+\tan^2\theta =\sec^2\theta$, we get, $\sec\theta=\sqrt{1+\frac{1}{3}}=\frac{2}{\sqrt3}$ So, $\frac{r\tan \theta +\sec \theta}{r\sec \theta +\tan\theta} = \frac{\frac{2}{\sqrt3}+\frac{2}{\sqrt3}}{\frac{4}{\sqrt3}+\frac{1}{\sqrt3}} =\frac{4}{5}$ Hence, the correct answer is $\frac{4}{5}$.
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