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  4. If cos (2 θ2)=( cos (θ1+θ3)/ cos (θ1-θ3)), then

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Q. If $\cos \left(2 \theta_2\right)=\frac{\cos \left(\theta_1+\theta_3\right)}{\cos \left(\theta_1-\theta_3\right)}$, then

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Solution:

$\cos \left(2 \theta_2\right)=\frac{\cos \left(\theta_1+\theta_3\right)}{\cos \left(\theta_1-\theta_3\right)} \Rightarrow \frac{\cos \left(2 \theta_2\right)}{1}=\frac{\cos \left(\theta_1+\theta_3\right)}{\cos \left(\theta_1-\theta_3\right)}$ (Using dividendo and componendo)
$\Rightarrow \frac{\cos \left(2 \theta_2\right)-1}{\cos \left(2 \theta_2\right)+1}=\frac{-2 \sin \theta_1 \sin \theta_3}{2 \cos \theta_1 \cos \theta_3} $
$\Rightarrow \tan ^2 \theta_2=\tan \theta_1 \tan \theta_3 \Rightarrow \tan \theta_1, \tan \theta_2, \tan \theta_3 \text { are in G.P.Ans.}$