If cos(θ1+θ2)cos(θ1−θ2)+cos(θ3−θ4)cos(θ3+θ4)=0 then cotθ1,cotθ2,cotθ3,cotθ4=

cos(θ1+θ2)cos(θ1−θ2)=−cos(θ3-θ4)cos(θ3+θ4) ⇒−2sinθ1sinθ22cosθ1cosθ2=−2cosθ3cosθ4−2sinθ3sinθ4 ⇒−tanθ1tanθ2=cotθ3cotθ4⇒cotθ1cotθ2cotθ3cotθ4=−1

If cos(θ1+θ2)cos(θ1−θ2)+cos(θ3−θ4)cos(θ3+θ4)=0 then cotθ1,cotθ2,cotθ3,cotθ4=

cos(θ1+θ2)cos(θ1−θ2)=−cos(θ3-θ4)cos(θ3+θ4) ⇒−2sinθ1sinθ22cosθ1cosθ2=−2cosθ3cosθ4−2sinθ3sinθ4 ⇒−tanθ1tanθ2=cotθ3cotθ4⇒cotθ1cotθ2cotθ3cotθ4=−1

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If $\frac{\cos (θ_{1} + θ_{2})}{\cos (θ_{1} - θ_{2})} + \frac{\cos (θ_{3} - θ_{4})}{\cos (θ_{3} + θ_{4})} = 0$ then $\cot θ_{1}, \cot θ_{2}, \cot θ_{3}, \cot θ_{4} =$

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Detailed Solution

$\frac{\cos (θ_{1} + θ_{2})}{\cos (θ_{1} - θ_{2})} = \frac{- \cos (θ_{3} - θ_{4})}{\cos (θ_{3} + θ_{4})} \Rightarrow \frac{- 2 \sin θ_{1} \sin θ_{2}}{2 \cos θ_{1} \cos θ_{2}} = \frac{- 2 \cos θ_{3} \cos θ_{4}}{- 2 \sin θ_{3} \sin θ_{4}} \begin{array}{l} \Rightarrow - \tan θ_{1} \tan θ_{2} = \cot θ_{3} \cot θ_{4} \\ \Rightarrow \cot θ_{1} \cot θ_{2} \cot θ_{3} \cot θ_{4} = - 1 \end{array}$