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If $3\tan \theta\tan\phi=1$ ,prove that $2\cos (\theta+\phi)=\cos (\theta-\phi)$

1 Answer

Toolbox:
  • $\cot\theta=\large\frac{\cos \theta}{\sin \theta}$
  • $\cos(\theta+\phi)=\cos \theta\cos\phi-\sin \theta\sin \phi$
  • $\cos(\theta-\phi)=\cos \theta\cos\phi+\sin \theta\sin \phi$
Given :
$3\tan \theta\tan\phi=1$ or $\cot \theta\cot\phi=3$
Or $\large\frac{\cos \theta.\cos \phi}{\sin \theta.\sin \phi}=\frac{3}{3}$
By componendo or dividendo,we have
$\large\frac{\cos \theta.\cos \phi+\sin\theta\sin\phi}{\cos\theta\cos\phi-\sin \theta.\sin \phi}=\frac{3+1}{3-1}$
$\large\frac{\cos(\theta-\phi)}{\cos(\theta+\phi)}$$=2$
$2\cos (\theta+\phi)=\cos (\theta-\phi)$
Hence proved
answered May 21, 2014 by sreemathi.v
 

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