Question

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

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

Answer: (B)

$1$

We have

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

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

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

$=\frac{\cos \theta sin3\theta }{\cos \theta \sin 3\theta -\cos 3\theta \sin \theta }+\frac{\sin \theta \cos 3\theta }{\sin \theta \cos 3\theta -\sin 3\theta \cos \theta }$

$=\frac{\cos \theta sin3\theta }{\cos \theta \sin 3\theta -\cos 3\theta \sin \theta }-\frac{\sin \theta \cos 3\theta }{\sin 3\theta \cos \theta -\sin \theta \cos 3\theta }$

$=\frac{\cos \theta sin3\theta -\sin \theta \cos 3\theta }{\cos \theta \sin 3\theta -\cos 3\theta \sin \theta }$

$=1$

Hence, the value is $1$.