Question

$\frac{\sin \theta -2{\sin }^3\theta }{2{\cos }^3\theta -\cos \theta }=\cot \theta$

• A. True
• B. False

Answer 1

The correct option is B False

$\frac{\sin \theta -2{\sin }^3\theta }{2{\cos }^3\theta -\cos \theta }$

$=\frac{\sin \theta (1-2{\sin }^2\theta )}{\cos \theta (2{\cos }^2\theta -1)}$

$=\frac{\sin \theta \left(\cos 2\theta \right)}{\cos \theta \left(\cos 2\theta \right)}$ since $1-2{\sin }^2\theta =2{\cos }^2\theta -1=\cos 2\theta$

$=\frac{\sin \theta }{\cos \theta }$

$=\tan \theta$

Hence the given statement is false.