Question

$\frac{\cot \theta }{\cot \theta -\cot 3\theta }+\frac{\tan \theta }{\tan \theta -\tan 3\theta }=$

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is B $1$

$\frac{\cot \theta }{\cot \theta -\cot 3\theta }+\frac{\tan \theta }{\tan \theta -\tan 3\theta }$

$=\frac{\frac{\cos \theta }{\sin \theta }}{\frac{\cos \theta }{\sin \theta }-\frac{\cos 3\theta }{\sin 3\theta }}+\frac{\frac{\cos \theta }{\sin \theta }}{\frac{\cos \theta }{\sin \theta }-\frac{\cos 3\theta }{\sin 3\theta }}$

$=\frac{\frac{\cos \theta }{\sin \theta }}{\frac{\sin 3\theta \times \cos \theta -\sin \theta \times \cos 3\theta }{\sin \theta \times \sin 3\theta }}+\frac{\frac{\sin \theta }{\cos \theta }}{\frac{\sin \theta \times \cos 3\theta -\sin 3\theta \times \cos \theta }{\cos \theta \times \cos 3\theta }}$

$=\frac{\frac{\cos \theta }{\sin \theta }}{\frac{\sin (3\theta -\theta )}{\sin \theta \times \sin 3\theta }}+\frac{\frac{\sin \theta }{\cos \theta }}{\frac{\sin (\theta -3\theta )}{\cos \theta \times \cos 3\theta }}$

$=\frac{\cos \theta \times \sin 3\theta }{\sin 2\theta }+\frac{\sin \theta \times \cos 3\theta }{\sin (-2\theta )}$

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

$=\frac{\sin (3\theta -\theta )}{\sin 2\theta }=\frac{\sin 2\theta }{\sin 2\theta }=1$

Hence, option (b) is correct.