cot θcot θ−cot 3θ+tan θtan θ−tan 3θis equal to
Question:

$\frac{\cot \theta}{\cot \theta-\cot 3 \theta}+\frac{\tan \theta}{\tan \theta-\tan 3 \theta}$ is equal to

(a) 0
(b) 1
(c) −1
(d) 2

Solution:

The given expression is $\frac{\cot \theta}{\cot \theta-\cot 3 \theta}+\frac{\tan \theta}{\tan \theta-\tan 3 \theta}$.

Simplifying the given expression, we have

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

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

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

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

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

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

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

$=1$

Therefore, the correct option is (b).

 

Administrator

Leave a comment

Please enter comment.
Please enter your name.