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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