Maths-
General
Easy

Question

If equals fraction numerator 2 pi over denominator 7 end fraction,then tan alpha tan invisible function application 2 alpha plus tan invisible function application 2 alpha tan invisible function application 4 alpha plus tan invisible function application 4 alpha tan invisible function application alpha equals

  1. negative 1
  2. negative 3
  3. negative 5
  4. negative 7

hintHint:

In this question, we have to find the value of tanalpha tan invisible function application 2 alpha plus tan invisible function application 2 alpha tan invisible function application 4 alpha plus tan invisible function application 4 alpha tan invisible function application alpha
, if alphaequals fraction numerator 2 pi over denominator 7 end fraction. For this we will solve the function using common trigonometric identity and simplify it and later substitute the value of alpha in the equation.

The correct answer is: negative 7


    tan space alpha. tan space 2 alpha plus tan space 2 alpha. tan space 4 alpha plus tan space 4 alpha. tan space alpha
equals fraction numerator sin space alpha over denominator cos space alpha end fraction. fraction numerator sin space 2 alpha over denominator cos space 2 alpha end fraction plus fraction numerator sin space 2 alpha over denominator cos space 2 alpha end fraction. fraction numerator sin space 4 alpha over denominator cos space 4 alpha end fraction plus fraction numerator sin space 4 alpha over denominator cos space 4 alpha end fraction. fraction numerator sin space alpha over denominator cos space alpha end fraction
equals fraction numerator sin space alpha. sin space 2 alpha. cos space 4 alpha plus cos space alpha. sin space 2 alpha. sin space 4 alpha plus sin space alpha. cos space 2 alpha. sin space 4 alpha over denominator cos space alpha. cos space 2 alpha. cos space 4 alpha end fraction
equals 1 minus open parentheses 1 minus fraction numerator sin space alpha. sin space 2 alpha. cos space 4 alpha plus cos space alpha. sin space 2 alpha. sin space 4 alpha plus sin space alpha. cos space 2 alpha. sin space 4 alpha over denominator cos space alpha. cos space 2 alpha. cos space 4 alpha end fraction close parentheses
equals 1 minus open parentheses fraction numerator cos space alpha. cos space 2 alpha. cos space 4 alpha plus sin space alpha. sin space 2 alpha. cos space 4 alpha plus cos space alpha. sin space 2 alpha. sin space 4 alpha plus sin space alpha. cos space 2 alpha. sin space 4 alpha over denominator cos space alpha. cos space 2 alpha. cos space 4 alpha end fraction close parentheses
equals 1 minus fraction numerator cos space left parenthesis alpha plus 2 alpha plus 4 alpha right parenthesis over denominator cos space alpha. cos space 2 alpha. cos space 4 alpha end fraction
equals 1 minus fraction numerator cos space 7 alpha over denominator begin display style fraction numerator sin space 8 alpha over denominator 8 space sin space alpha end fraction end style end fraction
equals 1 minus fraction numerator 8. sin space alpha. cos space 7 alpha over denominator sin space 8 alpha end fraction
S u b s t i t u t i n g space t h e space v a l u e space o f space alpha equals fraction numerator 2 pi over denominator 7 end fraction comma space
equals 1 minus fraction numerator 8. sin space open parentheses fraction numerator 2 pi over denominator 7 end fraction close parentheses. cos space 2 pi over denominator sin space open parentheses fraction numerator 2 pi over denominator 7 end fraction plus 2 pi close parentheses end fraction
equals 1 minus 8
equals negative 7

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