Question

If $ \cos (\theta +\phi )=m\cos (\theta -\phi ), $ then $ \tan \theta $ is equal to:

• A. $ [(1+m)/(1-m)]\tan \phi $
• B. $ [(1-m)/(1+m)]\tan \phi $
• C. $ [(1-m)/(1+m)]cot\phi $
• D. $ [(1+m)/(1-m)]sec\phi $

Answer 1

Answer: (C)

We have $\cos (\theta+\phi)=m \cos(\theta-\phi)$
$\Rightarrow \cos \theta \cos \phi-\sin \theta \sin \phi=m(\cos \theta \cos \phi+\sin \theta \sin \phi)$
$\Rightarrow \cos \theta \cos \phi(1-m)=\sin \theta \sin \phi(1+m)$
$\Rightarrow \tan \theta=\left(\frac{1-m}{1+m}\right) \cot \phi$