If cos(θ1+θ2)cos(θ1−θ2)+cos(θ3−θ4)cos(θ3+θ4)=0,then tanθ1tanθ2tanθ3tanθ4 is equal to

cos(θ1+θ2)cos(θ1−θ2)+cos(θ3−θ4)cos(θ3+θ4)=0⇒cosθ1cosθ2−sinθ1sinθ2cosθ1cosθ2+sinθ1sinθ2+cosθ3cosθ4+sinθ3sinθ4cosθ3cosθ4−sinθ3sinθ4=0⇒1−tanθ1tanθ21+tanθ1tanθ2+1+tanθ3tanθ41−tanθ3tanθ4=0⇒1+tanθ1tanθ2+tanθ3tanθ4+tanθ1tanθ2tanθ3tanθ4+1−tanθ3tanθ4−tanθ1tanθ2+tanθ1tanθ2tanθ3tanθ4=0⇒2+2tanθ1tanθ2tanθ3tanθ4=0⇒tanθ1tanθ2tanθ3tanθ4=−1

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If { (_{1} - _{2})}{ (_{1} + _{2})} + { (_{3} + _{4})}{ (_{3

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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 $tan θ_{1} tan θ_{2} tan θ_{3} tan θ_{4}$ is equal to

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Noah Discussed

\frac{cos ( θ _{1} - θ _{2} )}{cos ( θ _{1} + θ _{2} )} + \frac{cos ( θ _{3} + θ _{4} )}{cos ( θ _{3} - θ _{4} )} = 0

,
then

tan θ_{1} tan θ_{2} tan θ_{3} tan θ_{4}

is equal to

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Text solutionVerified

\frac{cos ( θ _{1} - θ _{2} )}{cos ( θ _{1} + θ _{2} )} + \frac{cos ( θ _{3} + θ _{4} )}{cos ( θ _{3} - θ _{4} )} = 0

\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{cos θ _{3} cos θ _{4} - sin θ _{3} sin θ _{4}}{cos θ _{3} cos θ _{4} + sin θ _{3} sin θ _{4}} = 0

\Rightarrow \frac{1 + tan θ _{1} tan θ _{2}}{1 - tan θ _{1} tan θ _{2}} + \frac{1 - tan θ _{3} tan θ _{4}}{1 + tan θ _{3} tan θ _{4}} = 0

\Rightarrow 1 + tan θ_{1} tan θ_{2} + tan θ_{3} tan θ_{4} + tan θ_{1} tan θ_{2} tan θ_{3} tan θ_{4} + 1 - tan θ_{3} tan θ_{4} - tan θ_{1} tan θ_{2} + tan θ_{1} tan θ_{2} tan θ_{3} tan θ_{4} = 0

\Rightarrow 2 + 2 tan θ_{1} tan θ_{2} tan θ_{3} tan θ_{4} = 0

\Rightarrow tan θ_{1} tan θ_{2} tan θ_{3} tan θ_{4} = - 1

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Question Text	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 $tan θ_{1} tan θ_{2} tan θ_{3} tan θ_{4}$ is equal to
Answer Type	Text solution:1
Upvotes	150

If cos(θ1​+θ2​)cos(θ1​−θ2​)​+cos(θ3​−θ4​)cos(θ3​+θ4​)​=0,then tanθ1​tanθ2​tanθ3​tanθ4​ is equal to

Text solutionVerified

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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 $tan θ_{1} tan θ_{2} tan θ_{3} tan θ_{4}$ is equal to