Question

$\underset{\theta \to 0}{lim}\frac{4\theta (\tan \theta -\sin \theta )}{{(1-\cos 2\theta )}^2}$ is equal to

• A. $-\frac{1}{\sqrt{2}}$
• B. $\frac{1}{2}$
• C. $1$
• D. $2$

Answer 1

The correct option is B $\frac{1}{2}$

$\underset{\theta \to 0}{lim}\frac{4\theta (\tan \theta -\sin \theta )}{{(1-\cos 2\theta )}^2}=\underset{\theta \to 0}{lim}\frac{4\theta \sin \theta (1-\cos \theta )}{{\left(2{\sin }^2\theta \right)}^2.\cos \theta }$

$=\underset{\theta \to 0}{lim}\frac{\theta .(1-\cos \theta )}{\sin \theta .{\sin }^2\theta .\cos \theta }=\underset{\theta \to 0}{lim}\frac{\theta }{\sin \theta }\times \underset{\theta \to 0}{lim}\frac{(1-\cos \theta )(1+\cos \theta )}{{\sin }^2\theta .\cos \theta .(1+\cos \theta )}$

$=\frac{1}{{lim}_{\theta \to 0}\left(\frac{\sin \theta }{\theta }\right)}\times \underset{\theta \to 0}{lim}\frac{(1-{\cos }^2\theta )}{{\sin }^2\theta \cos \theta (1+\cos \theta )}$

$=\frac{1}{1}\times \underset{\theta \to 0}{lim}\frac{{\sin }^2\theta }{{\sin }^2\theta .\cos \theta (1+\cos \theta )}=\underset{\theta \to 0}{lim}\frac{1}{\cos \theta (1+\cos \theta )}=\frac{1}{2}$