If tanθ1=λcotθ2 then cos(θ1−θ2)cos(θ1+θ2)=

tanθ1tanθ2=λsinθ1sinθ2cosθ1cosθ2=λ1⇒cosθ1cosθ2+sinθ1sinθ2cosθ1cosθ2−sinθ1sinθ2=1+λ1−λcos(θ1−θ2)cos(θ1+θ2)=1+λ1−λ

If tanθ1=λcotθ2 then cos(θ1−θ2)cos(θ1+θ2)=

tanθ1tanθ2=λsinθ1sinθ2cosθ1cosθ2=λ1⇒cosθ1cosθ2+sinθ1sinθ2cosθ1cosθ2−sinθ1sinθ2=1+λ1−λcos(θ1−θ2)cos(θ1+θ2)=1+λ1−λ

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If $\tan θ_{1} = λ \cot θ_{2}$ then $\frac{\cos (θ_{1} - θ_{2})}{\cos (θ_{1} + θ_{2})}$ =

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$\frac{1 + λ}{1 - λ}$

$\frac{1 - λ}{1 + λ}$

$\frac{λ + 1}{λ - 1}$

$\frac{λ - 1}{λ + 1}$

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Detailed Solution

$\tan θ_{1} \tan θ_{2} = λ$

$\frac{\sin θ_{1} \sin θ_{2}}{\cos θ_{1} \cos θ_{2}} = \frac{λ}{1}$

$\Rightarrow \frac{\cos θ_{1} \cos θ_{2} + \sin θ_{1} \sin θ_{2}}{\cos θ_{1} \cos θ_{2} - \sin θ_{1} \sin θ_{2}} = \frac{1 + λ}{1 - λ}$

$\frac{\cos (θ_{1} - θ_{2})}{\cos (θ_{1} + θ_{2})} = \frac{1 + λ}{1 - λ}$

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If tanθ1=λcotθ2 then cos(θ1−θ2)cos(θ1+θ2)=

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If $\tan θ_{1} = λ \cot θ_{2}$ then $\frac{\cos (θ_{1} - θ_{2})}{\cos (θ_{1} + θ_{2})}$ =