A) \[2\]
B) \[-\frac{1}{2}\]
C) \[\frac{1}{2}\]
D) \[-2\]
Correct Answer: A
Solution :
\[\frac{{{(2{{\sin }^{2}}x)}^{2}}}{x\left[ x\left( x-\frac{{{x}^{3}}}{3}+\frac{2{{x}^{5}}}{15}...... \right)-\left( 2x-\frac{8{{x}^{3}}}{3}...... \right) \right]}\] \[=\frac{4{{\sin }^{4}}x}{{{x}^{4}}\left[ -\frac{2}{3}+\frac{8}{3} \right]}=\frac{4}{2}=2\]
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