# If $\cos(\theta+\phi)=m\cos(\theta-\phi)$ then prove that $\tan\theta=\large\frac{1-m}{1+m}$$\cot\phi ## 1 Answer Toolbox: • \cos a+\cos b=2\cos(\large\frac{a+b}{2})$$\cos(\large\frac{a-b}{2})$
• $\cos a-\cos b=-2\sin(\large\frac{a+b}{2})$$\sin (\large\frac{a-b}{2}) Given \cos(\theta+\phi)=m\cos(\theta-\phi) \large\frac{\cos(\theta+\phi)}{\cos(\theta-\phi)}=\frac{m}{1} Applying componendo and dividendo \large\frac{cos(\theta+\phi)+\cos(\theta-\phi)}{cos(\theta+\phi)-\cos(\theta-\phi)}=\frac{m+1}{m-1} \large\frac{2\cos(\Large\frac{\theta+\phi+\theta-\phi)}{2}.\cos(\Large\frac{\theta+\phi-\theta+\phi}{2})}{-2\sin(\Large\frac{\theta+\phi+\theta-\phi)}{2}.\sin(\Large\frac{\theta+\phi-\theta+\phi}{2})}=\frac{m+1}{m-1} \large\frac{\cos \theta.\cos\phi}{-\sin \theta.\sin \theta}=\frac{m+1}{m-1} -\cot \theta.\cot\phi=\large\frac{m+1}{m-1} -\cot \theta=\large\frac{m+1}{m-1}\frac{1}{\cot \phi} -\large\frac{1}{\tan \theta}=\frac{m+1}{m-1}$$\times \large\frac{1}{\cot \phi}$
$-\cot \phi.\large\frac{m-1}{1+m}=$$\tan \theta \tan\theta=\large\frac{1-m}{1+m}$$\cot\phi$
Hence proved.