If cos (θ + ϕ) = m cos (θ – ϕ),


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]


According to the question,

$\cos (\theta+\phi)=m \cos (\theta-\phi)$

$\because \cos (\theta+\phi)=\mathrm{m} \cos (\theta-\phi)$

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

Applying componendo - dividend, we get,

$\Rightarrow \frac{\cos (\theta-\phi)+\cos (\theta+\phi)}{\cos (\theta-\phi)-\cos (\theta+\phi)}=\frac{1+\mathrm{m}}{1-\mathrm{m}}$

From transformation formula, we know that,

$\cos (A+B)+\cos (A-B)=2 \cos A \cos B$


$\cos (A-B)-\cos (A+B)=2 \sin A \sin B$

$\Rightarrow \frac{2 \cos \theta \cos \phi}{2 \sin \theta \sin \phi}=\frac{1+m}{1-m}$

Since, $(\cos \theta) /(\sin \theta)=\cot \theta$

$\Rightarrow \cot \theta \cot \phi=\frac{1+m}{1-m}$

$\Rightarrow\left(\frac{1-\mathrm{m}}{1+\mathrm{m}}\right) \cot \phi=\frac{1}{\cot \theta}$

$\Rightarrow \tan \theta=\left(\frac{1-\mathrm{m}}{1+\mathrm{m}}\right) \cot \phi$

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