| 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 |
A) 5/4
B) 4/6
C) \[\sqrt{3}/2\]
D) ½
Correct Answer: B
Solution :
| \[{{(r\cos \theta -\sqrt{3})}^{2}}=0\] |
| \[\Rightarrow \] \[(r\cos \theta -\sqrt{3})=0\] |
| \[\Rightarrow \] \[r\cos \theta =\sqrt{3}\] (i) |
| \[\Rightarrow \] \[{{(r\sin \theta -1)}^{2}}=0\] |
| \[\Rightarrow \] \[(r\sin \theta -1)=0\]\[\Rightarrow \]\[r\sin \theta =1\] (ii) |
| On squaring Eqs. (i) and (ii) and adding, |
| \[{{r}^{2}}({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )=3+1=4\]\[\Rightarrow \]\[r=2\] |
| \[\cos \theta =\frac{\sqrt{3}}{2}\]\[\Rightarrow \]\[\theta =\frac{\pi }{6}\] |
| \[\Rightarrow \] \[\sin \theta =\frac{1}{2}\]\[\Rightarrow \]\[\theta =\frac{\pi }{6}\] |
| \[\frac{r\tan \theta +\sec \theta }{r\sec \theta +\tan \theta }=\frac{2\tan \pi /6+\sec \pi /6}{2\sec \pi /6+\tan \pi /6}\] |
| \[=\,\,\,\frac{2\times \frac{1}{\sqrt{3}}+\frac{2}{\sqrt{3}}}{2\times \frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}}}\,\,=\,\,\frac{\frac{4}{\sqrt{3}}}{\frac{5}{\sqrt{3}}}\,\,=\,\,\frac{4}{5}\] |
You need to login to perform this action.
You will be redirected in
3 sec
