| If \[2x=\sec A\] and \[\frac{2}{x}=\tan A,\] then \[2\left( {{x}^{2}}-\frac{1}{{{x}^{2}}} \right)\]is equal to |
A) \[\frac{1}{2}\]
B) \[\frac{1}{4}\]
C) \[\frac{1}{8}\]
D) \[\frac{1}{16}\]
Correct Answer: A
Solution :
| \[\left( 2x+\frac{2}{x} \right)\left( 2x-\frac{2}{x} \right)=(\sec A+\tan A)(\sec A-\tan A)\]\[\Rightarrow \]\[4\left( {{x}^{2}}-\frac{1}{{{x}^{2}}} \right)=\left( \frac{1+\sin A}{\cos A} \right)\left( \frac{1-\sin A}{\cos A} \right)=\frac{{{\cos }^{2}}A}{{{\cos }^{2}}A}=1\] |
| \[\Rightarrow \] \[2\left( {{x}^{2}}-\frac{1}{{{x}^{2}}} \right)=\frac{1}{2}\] |
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