Question

cot theta upon cot theta minus Cot 3 theta + tan theta upon tan theta minus tan 3 theta is equal to

Answer 1

Answer:

Step-by-step explanation:i have written the answers with the proper steps. Refer below

Answer 2

The value of $\frac{\cot \theta}{\cot \theta-\cot 3 \theta}+\frac{\tan \theta}{\tan \theta-\tan 3 \theta}$ is 1

Given:

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

To find:

The value of $\frac{\cot \theta}{\cot \theta-\cot 3 \theta}+\frac{\tan \theta}{\tan \theta-\tan 3 \theta}$

Solution:

To simplify, $\frac{\cot \theta}{\cot \theta-\cot 3 \theta}+\frac{\tan \theta}{\tan \theta-\tan 3 \theta}$

Take LCM, the equation becomes,  

$=\frac{[\cot \theta \times(\tan \theta-\tan 3 \theta)][\tan \theta \times(\cot \theta-\cot 3 \theta]}{(\cot \theta-\cot 3 \theta) \times(\tan \theta-\tan 3 \theta)}$

$=\frac{[\cot \theta \times(\tan \theta-\tan 3 \theta)][\tan \theta \times(\cot \theta-\cot 3 \theta]}{(\cot \theta \times \tan \theta)-(\cot 3 \theta \times \tan \theta)+(\cot \theta \times \tan \theta)-(\tan 3 \theta \times \cot \theta)}$

Since $\cot \theta \times \tan \theta=1$

$=\frac{1-(\cot \theta \times \tan 3 \theta)+1-(\tan \theta \times \cot 3 \theta)}{(1)-(\cot 3 \theta \times \tan \theta)+(1)-(\tan 3 \theta \times \cot \theta)}$

$=\frac{2-(\cot \theta \times \tan 3 \theta)-(\tan \theta \times \cot 3 \theta)}{(2)-(\cot 3 \theta \times \tan \theta)-(\tan 3 \theta \times \cot \theta)}$

Since numerator and denominator are equal, $\frac{\cot \theta}{\cot \theta-\cot 3 \theta}+\frac{\tan \theta}{\tan \theta-\tan 3 \theta}=1$