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I`d like you guys to please solve question number 6. Thanks!

Ritika Rakesh , 7 Years ago
Grade 12
anser 3 Answers
Sayam

Last Activity: 7 Years ago

Answer is C
as cot\theta-cot3\theta
=cos\theta/sin\theta-cos3\theta/sin3\theta
=2 cos\thetasin3\theta-2cos3\thetasin\theta/2sin\thetasin3\theta  …. taking LCM
=sin4\theta+sin2\theta-sin4\theta+sin2\theta/2sin3\thetasin\theta
=2cos\thetacosec3\theta
Hopes this clears your doubt. Thanks

Sarasij Basu Mallick

Last Activity: 7 Years ago

 \theta-cot3 \thetacot
=(cos\theta /sin\theta )-(cos3\theta /sin3\theta )
=((sin3\theta cos\theta -cos3\theta sin\theta )/(sin\theta sin3\theta ))
=((sin2\theta)/(sin\thetasin3\theta ))
=((2sin\theta cos\theta )/(sin\theta sin3\theta ))
=2cos\theta cosec3\theta --------------c 
 
 

Sarasij Basu Mallick

Last Activity: 7 Years ago

---------SORRY SORRY SOMETHING WENT WRONG IN TYPING----------
cot\theta-cot3\theta
=(cos\theta /sin\theta )-(cos3\theta /sin3\theta )
=((sin3\theta cos\theta -cos3\theta sin\theta )/(sin\theta sin3\theta ))
=((sin2\theta)/(sin\thetasin3\theta ))
=((2sin\theta cos\theta )/(sin\theta sin3\theta ))
=2cos\theta cosec3\theta --------------c 
 
 

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