Question

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

• A. $\left(\frac{1+m}{1-m}\right)\tan \varphi$
• B. $\left(\frac{1-m}{1+m}\right)\tan \varphi$
• C. $\left(\frac{1+m}{1-m}\right)\cot \varphi$
• D. $\left(\frac{1-m}{1+m}\right)\cot \varphi$

Answer 1

Answer: (D)

$\left(\frac{1-m}{1+m}\right)\cot \varphi$

$\frac{\cos (\theta +\varphi )}{\cos (\theta -\varphi )}=m$

Apply componendo and dividendo

$\frac{\cos (\theta +\varphi )+\cos (\theta -\varphi )}{\cos (\theta +\varphi )-\cos (\theta -\varphi )}=\frac{m+1}{m-1}$

$\frac{\cos (\theta +\varphi )+\cos (\theta -\varphi )}{-(\cos (\theta -\varphi )-\cos (\theta +\varphi ))}=\frac{m+1}{m-1}$

$\frac{2\cos \left(\theta \right)\cdot \cos \left(\varphi \right)}{-2\sin \left(\theta \right)\cdot \sin \left(\varphi \right)}=\frac{1+m}{m-1}$

$\frac{2\cos \left(\theta \right)\cdot \cos \left(\varphi \right)}{2\sin \left(\theta \right)\cdot \sin \left(\varphi \right)}=\frac{1+m}{1-m}$

$\cot \theta \cdot \cot \varphi =\frac{1+m}{1-m}$

$\tan \theta =\left(\frac{1-m}{1+m}\right)\cdot \cot \varphi$