If cos-1-2-cos-1-2-cos-3-4-cos-3-4-0- then tan-1tan-2tan-3tan-4-

The correct option is C 1Given, cos(θ _1 θ _2)/cos(θ _1+θ _2)=cos(θ _3 θ _4)/cos(θ _3 θ _4)⇒cos(θ _1 θ _2)+cos(θ _1+θ _2)/cos(θ _1 θ _2) cos(θ _1+θ _2)= [ c ...

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Byju's Answer

Standard XI

Mathematics

Trigonometric Ratios of Multiples of an Angle

If cosθ1-θ2/c...

Question

If $\frac{c o s (θ_{1} - θ_{2})}{c o s (θ_{1} - θ_{2})} + \frac{c o s (θ_{3} + θ_{4})}{c o s (θ_{3} - θ_{4})} = 0, then tan θ_{1} t a n θ_{2} t a n θ_{3} t a n θ_{4} =$

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Solution

The correct option is C -1
Given, $\frac{c o s (θ_{1} - θ_{2})}{c o s (θ_{1} + θ_{2})} = \frac{c o s (θ_{3} - θ_{4})}{c o s (θ_{3} - θ_{4})} \Rightarrow \frac{c o s (θ_{1} - θ_{2}) + c o s (θ_{1} + θ_{2})}{c o s (θ_{1} - θ_{2}) - c o s (θ_{1} + θ_{2})} = [\frac{- c o s (θ_{3} + θ_{4}) + c o s (θ_{3} - θ_{4})}{- c o s (θ_{3} + θ_{4}) - c o s (θ_{3} - θ_{4})}]$

$\Rightarrow \frac{2 c o s θ_{1} c o s θ_{2}}{2 s i n θ_{1} s i n θ_{2})} = \frac{2 s i n θ_{3} s i n θ_{4}}{- 2 c o s θ_{3} c o s θ_{4}} \Rightarrow c o t θ_{1} c o t θ_{2} = - t a n θ_{3} t a n θ_{4} \Rightarrow t a n θ_{1} t a n θ_{2} t a n θ_{4} = - 1$

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Similar questions

Q. Let A and B denote the statements
A:

c o s θ_{1} + c o s θ_{2} + c o s θ_{3} = 0

s i n θ_{1} + s i n θ_{2} + s i n θ_{3} = 0.

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sin θ_{1} + sin θ_{2} + sin θ_{3} = 3, 0^{o} < θ_{1}, θ_{2}, θ_{3} < 90^{o}

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If cos(θ1−θ2)cos(θ1−θ2)+cos(θ3+θ4)cos(θ3−θ4)= 0, then tan θ1 tan θ2 tan θ3 tan θ4=

If $\frac{c o s (θ_{1} - θ_{2})}{c o s (θ_{1} - θ_{2})} + \frac{c o s (θ_{3} + θ_{4})}{c o s (θ_{3} - θ_{4})} = 0, then tan θ_{1} t a n θ_{2} t a n θ_{3} t a n θ_{4} =$