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  • If cos (θ + ϕ) = m cos (θ – ϕ), then prove that tan θ = ((1 – m)/(1 + m)) cot ϕ [Hint: Express cos (θ + ϕ)/ cos (θ – ϕ) = m/l and apply Componendo and Dividendo]

If \cos ( \theta + \phi ) = m \cos ( \theta - \phi ), then prove that \tan \theta = ((1 - m)/(1 + m)) cot \phi
[Hint: Express \cos ( \theta + \phi )/ \cos ( \theta - \phi ) = m/1 and apply Componendo and Dividendo]

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\\ \text{Given that}\cos \left( \theta + \phi \right) =m\cos \left( \theta - \phi \right) ~~ \\\\ ~~~\frac{\cos \left( \phi + \theta \right) }{\cos \left( \theta - \phi \right) }=\frac{m}{1} \\\\

\text{~~Using~componendo~and dividend rule, }\frac{\cos \left( \phi + \theta \right) +\cos \left( \theta - \phi \right) }{\cos \left( \phi + \theta \right) - \cos \left( \theta - \phi \right) } =\frac{m+1}{m - 1}~ \\\\
 \\ ~~ \frac{\cos \theta \cos \phi }{ - \sin \theta \sin \phi }=\frac{m+1}{m - 1}~~ \\\\ - cot \theta cot \phi =\frac{m+1}{m - 1}~ \\\\ ~\frac{ - cot \phi }{\tan \theta }=\frac{m+1}{m - 1} \\\\ ~~ \tan \theta \left( 1+m \right) = \left( 1 - m \right) cot \phi ~ \\\\

  \\ ~ \tan \theta =\frac{1 - m}{1+m}cot \phi \\\\

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