If tan θ1=k cot θ2 then cos(θ1−θ2)cos(θ1+θ2)=

Given that tanθ1.tanθ2=k.cos(θ1−θ2)cos(θ1+θ2)=cosθ1cosθ2+sinθ1sinθ2cosθ1cosθ2−sinθ1sinθ2=1+tanθ1tanθ21−tanθ1tanθ2=1+k1−k.

If tan θ1=k cot θ2 then cos(θ1−θ2)cos(θ1+θ2)=

Given that tanθ1.tanθ2=k.cos(θ1−θ2)cos(θ1+θ2)=cosθ1cosθ2+sinθ1sinθ2cosθ1cosθ2−sinθ1sinθ2=1+tanθ1tanθ21−tanθ1tanθ2=1+k1−k.

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$I f \tan θ_{1} = k \cot θ_{2} t h e n \frac{\cos (θ_{1} - θ_{2})}{\cos (θ_{1} + θ_{2})} =$

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$\frac{k + 1}{k - 1}$

$\frac{1 - k}{1 + k}$

$\frac{k - 1}{k + 1}$

$\frac{1 + k}{1 - k}$

answer is A.

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Detailed Solution

$\begin{array}{l} G i v e n t h a t \tan θ_{1} . \tan θ_{2} = k . \\ \frac{\cos (θ_{1} - θ_{2})}{\cos (θ_{1} + θ_{2})} = \frac{\cos θ_{1} \cos θ_{2} + \sin θ_{1} \sin θ_{2}}{\cos θ_{1} \cos θ_{2} - \sin θ_{1} \sin θ_{2}} = \frac{1 + \tan θ_{1} \tan θ_{2}}{1 - \tan θ_{1} \tan θ_{2}} = \frac{1 + k}{1 - k} . \end{array}$