Maths-
General
Easy

Question

If tan space theta subscript 1 comma tan space theta subscript 2 comma tan space theta subscript 3 and tan space theta subscript 4 are the roots of the equation x to the power of 4 minus x cubed sin space 2 beta plus x squared cos space 2 beta minus x cos space beta minus sin space beta equals 0  then tan space open parentheses theta subscript 1 plus theta subscript 2 plus theta subscript 3 plus theta subscript 4 close parentheses is equal to

  1. sin space beta
  2. cos space beta
  3. tan space beta
  4. c o t space beta

The correct answer is: c o t space beta


    x to the power of 4 minus x cubed sin space 2 beta plus x squared cos space 2 beta minus x cos space beta minus sin space beta equals 0
F o r space a b o v e space e q u a t i o n comma
tan space theta subscript 1 comma tan space theta subscript 2 comma tan space theta subscript 3 space a n d space tan theta subscript 4 space a r e space t h e space r o o t s

F r o m space t h e space a b o v e space e q u a t i o n comma space w e space g e t
sum from r equals 1 to 4 of tan theta subscript r space equals s i n 2 beta equals S subscript 1
sum from r equals 1 to 4 of tan theta subscript 1 tan space theta subscript 2 tan space theta subscript 3 space equals c o s 2 beta equals S subscript 2
sum from r equals 1 to 4 of tan theta subscript 1 tan space theta subscript 2 space equals c o s beta equals S subscript 3
sum from r equals 1 to 4 of tan theta subscript 1 tan space theta subscript 2 tan space theta subscript 3 tan theta subscript 4 space equals negative s i n beta equals S subscript 4

P u t t i n g space t h e space v a l u e s space i n space t h e space f o r m u l a comma
tan space open parentheses theta subscript 1 plus theta subscript 2 plus theta subscript 3 plus theta subscript 4 close parentheses
equals fraction numerator S subscript 1 minus S subscript 3 over denominator 1 minus S subscript 2 plus S subscript 4 end fraction
equals fraction numerator s i n 2 beta minus c o s beta over denominator 1 minus cos 2 beta plus open parentheses negative sin beta close parentheses end fraction
equals fraction numerator 2 sin beta cos beta minus cos beta over denominator 2 sin squared beta minus sin beta end fraction
equals fraction numerator cos beta open parentheses 2 sin beta minus 1 close parentheses over denominator sin beta open parentheses 2 sin beta minus 1 close parentheses end fraction
equals fraction numerator cos beta over denominator sin beta end fraction
equals c o t beta

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