A) \[1/\sqrt{2}\]
B) 1/2
C) 1
D) 2
Correct Answer: B
Solution :
\[\underset{\theta \to 0}{\mathop{\lim }}\,\frac{4\theta (\tan \theta -\sin \theta )}{{{(1-\cos 2\theta )}^{2}}}=\underset{\theta \to 0}{\mathop{\lim }}\,\frac{4\theta \sin \theta (1-\cos \theta )}{4{{\sin }^{4}}\theta \cos \theta }\] \[=\underset{\theta \to 0}{\mathop{\lim }}\,\left( \frac{\theta }{\sin \theta } \right)\frac{2{{\sin }^{2}}\theta /2}{{{\sin }^{2}}\theta \cos \theta }\] \[=\underset{\theta \to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}\theta /2}{(2\sin (\theta /2)\cos {{(\theta /2)}^{2}})}\frac{1}{\cos \theta }\] \[=\underset{\theta \to 0}{\mathop{\lim }}\,\frac{1}{2}\frac{1}{{{\cos }^{2}}(\theta /2).\cos \theta }=\frac{1}{2}\].
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