A) \[\frac{\sqrt{145}}{10}\]
B) \[\frac{\sqrt{145}}{11}\]
C) \[\frac{\sqrt{145}}{12}\]
D) \[\frac{\sqrt{146}}{12}\]
Correct Answer: C
Solution :
\[\cos {{\,}^{-1}}\,\left( \frac{2}{3x} \right)\,\,+\,\,\cos {{\,}^{-1}}\,\left( \frac{3}{4x} \right)\,\,=\,\,\frac{\pi }{2}\] \[\Rightarrow \,\,\,\cos {{\,}^{-1}}\,\left( \frac{2}{3x}\,\,.\,\frac{3}{4x}-\sqrt{1-\frac{4}{9{{x}^{2}}}}\,\sqrt{1-\frac{9}{16{{x}^{2}}}} \right)\,\,=\,\,\frac{\pi }{2}\] \[\Rightarrow \,\,\,\frac{1}{2{{x}^{2}}}\,\,=\,\,\sqrt{1-\frac{4}{9{{x}^{2}}}}\,\,\sqrt{1-\frac{9}{16{{x}^{2}}}}\] Square \[\Rightarrow \,\,\,\frac{1}{4{{x}^{2}}}\,\,=\,\,\left( 1-\frac{4}{9{{x}^{2}}} \right)\,\,\left( 1-\frac{9}{16{{x}^{2}}} \right)\] \[\Rightarrow \,\,\,\frac{1}{4{{x}^{2}}}\,\,=\,\,1\,-\,\frac{9}{16\,{{x}^{2}}}\,-\,\frac{4}{9{{x}^{2}}}\,+\,\frac{1}{4{{x}^{4}}}\] \[\Rightarrow \,\,\,\frac{9}{16{{x}^{2}}}\,\,+\,\,\frac{4}{9{{x}^{2}}}\,\,=\,\,1\] \[\frac{1}{{{x}^{2}}}\,\left( \frac{9}{16}\,\,+\,\,\frac{4}{9} \right)\,\,=\,\,1\] \[\Rightarrow \,\,\,\frac{81+64}{16\,\,\times \,\,9}\,\,=\,\,{{x}^{2}}\] \[\frac{145}{16\,\times 9}\,\,=\,\,{{x}^{2}}\]
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