A) \[\frac{3}{10}\]
B) \[\frac{-10}{3}\]
C) \[\frac{10}{3}\]
D) \[\frac{-20}{3}\]
Correct Answer: C
Solution :
| Let \[\cot \theta =x\] |
| and \[\sqrt{3}{{x}^{2}}-4x+\sqrt{3}=0\] |
| Here, \[a=\sqrt{3}\], \[b=-4\] and \[c=\sqrt{3}\] |
| \[X=\frac{-b\,\underline{+}\,\sqrt{D}}{2a}\] |
| [using quadratic formula] |
| \[\Rightarrow \,\,x=\frac{-\left( -4 \right)\underline{+}\,\sqrt{16-4.\sqrt{3}.\sqrt{3}}}{2\times \sqrt{3}}=\frac{4\underline{+}\sqrt{4}}{2\sqrt{3}}\] |
| \[\Rightarrow \,\,x=\frac{4+2}{2\sqrt{3}},\,\frac{4-2}{2\sqrt{3}}\] |
| \[\Rightarrow \,\,\cot \,\theta =\frac{6}{2\sqrt{3}},\,\frac{2}{2\sqrt{3}}\Leftrightarrow \cot \theta =\sqrt{3},\,\frac{1}{\sqrt{3}}\] |
| \[\Rightarrow \,\,\tan \theta =\frac{1}{\sqrt{3}},\,\sqrt{3}\] |
| \[\Rightarrow \,{{\cot }^{2}}\theta +{{\tan }^{2}}\theta ={{\left( \sqrt{3} \right)}^{2}}+{{\left( \frac{1}{\sqrt{3}} \right)}^{2}}\] |
| \[=3+\frac{1}{3}=\frac{10}{3}\] |
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