Home
Class 11
MATHS
sin ^(2) " (pi)/(6) + cos ^(2) "" (pi)/(...

`sin ^(2) " (pi)/(6) + cos ^(2) "" (pi)/(3) - tan ^(2) " (pi)/(4) =- 1/2`

Answer

Step by step text solution for sin ^(2) " (pi)/(6) + cos ^(2) "" (pi)/(3) - tan ^(2) " (pi)/(4) =- 1/2 by MATHS experts to help you in doubts & scoring excellent marks in Class 11 exams.

Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • TRIGONOMETRIC FUNCTIONS

    NCERT TAMIL|Exercise EXERCISE 3.4|9 Videos
  • TRIGONOMETRIC FUNCTIONS

    NCERT TAMIL|Exercise MISCELLANEOUS EXAMPLES|10 Videos
  • TRIGONOMETRIC FUNCTIONS

    NCERT TAMIL|Exercise EXERCISE 3.2|10 Videos
  • STRAIGHT LINES

    NCERT TAMIL|Exercise Miscellaneous Exercise on Chapter 10|24 Videos
  • TRIGONOMETRY

    NCERT TAMIL|Exercise EXERCISE 3.12|20 Videos

Similar Questions

Explore conceptually related problems

2 sin ^(2) "" (pi)/(6) + cosec ^(2) "" (7pi)/(6) cos ^(2) "" (pi)/(3) = 3/2

2 sin ^(2) "" (3pi)/(4) + 2 cos ^(2) "" (pi)/( 4) + 2 sec ^(2) "" (pi)/(3) =10

Knowledge Check

  • sin ^2 ""(2pi )/(7) + sin ^2"" (3pi )/(14) + sin ^2 ""(11 pi)/(14) + sin ^2 ""(5 pi)/(7) is

    A
    1
    B
    `3/2`
    C
    `2`
    D
    `3`
  • cos^2 ""(pi)/(8) + cos ^2 ""(3 pi)/(8) + cos ^2 ""(5 pi)/(8) + cos ^2 ""(7 pi)/(8) is

    A
    1
    B
    2
    C
    `-2`
    D
    `sqrt(2)`
  • Similar Questions

    Explore conceptually related problems

    cot ^(2) "" (pi)/(6) + cosec ""(5pi)/(6) + 3 tan ^(2) ""(pi)/(6) =6

    Show that sin^(2)""(pi)/(18) + sin^(2)""(pi)/(9) + sin^(2)""(7pi)/(18) + sin^(2)"" (4pi)/(9) = 2 .

    To find the sum sin^(2) ""(2pi)/(7) + sin^(2)""(4pi)/(7) +sin^(2)""(8pi)/(7) , we follow the following method. Put 7theta = 2npi , where n is any integer. Then " " sin 4 theta = sin( 2npi - 3theta) = - sin 3theta This means that sin theta takes the values 0, pm sin (2pi//7), pmsin(2pi//7), pm sin(4pi//7), and pm sin (8pi//7) . From Eq. (i), we now get " " 2 sin 2 theta cos 2theta = 4 sin^(3) theta - 3 sin theta or 4 sin theta cos theta (1-2 sin^(2) theta)= sin theta ( 4sin ^(2) theta -3) Rejecting the value sin theta =0 , we get " " 4 cos theta (1-2 sin^(2) theta ) = 4 sin ^(2) theta - 3 or 16 cos^(2) theta (1-2 sin^(2) theta)^(2) = ( 4sin ^(2) theta -3)^(2) or 16(1-sin^(2) theta) (1-4 sin^(2) theta + 4 sin ^(4) theta) " " = 16 sin ^(4) theta - 24 sin ^(2) theta +9 or " " 64 sin^(6) theta - 112 sin^(4) theta - 56 sin^(2) theta -7 =0 This is cubic in sin^(2) theta with the roots sin^(2)( 2pi//7), sin^(2) (4pi//7), and sin^(2)(8pi//7) . The sum of these roots is " " sin^(2)""(2pi)/(7) + sin^(2)""(4pi)/(7) + sin ^(2)""(8pi)/(7) = (112)/(64) = (7)/(4) . The value of (tan^(2)""(pi)/(7) + tan^(2)""(2pi)/(7) + tan^(2)""(3pi)/(7))xx (cot^(2)""(pi)/(7) + cot^(2)""(2pi)/(7) + cot^(2)""(3pi)/(7)) is

    To find the sum sin^(2) ""(2pi)/(7) + sin^(2)""(4pi)/(7) +sin^(2)""(8pi)/(7) , we follow the following method. Put 7theta = 2npi , where n is any integer. Then " " sin 4 theta = sin( 2npi - 3theta) = - sin 3theta This means that sin theta takes the values 0, pm sin (2pi//7), pmsin(2pi//7), pm sin(4pi//7), and pm sin (8pi//7) . From Eq. (i), we now get " " 2 sin 2 theta cos 2theta = 4 sin^(3) theta - 3 sin theta or 4 sin theta cos theta (1-2 sin^(2) theta)= sin theta ( 4sin ^(2) theta -3) Rejecting the value sin theta =0 , we get " " 4 cos theta (1-2 sin^(2) theta ) = 4 sin ^(2) theta - 3 or 16 cos^(2) theta (1-2 sin^(2) theta)^(2) = ( 4sin ^(2) theta -3)^(2) or 16(1-sin^(2) theta) (1-4 sin^(2) theta + 4 sin ^(4) theta) " " = 16 sin ^(4) theta - 24 sin ^(2) theta +9 or " " 64 sin^(6) theta - 112 sin^(4) theta - 56 sin^(2) theta -7 =0 This is cubic in sin^(2) theta with the roots sin^(2)( 2pi//7), sin^(2) (4pi//7), and sin^(2)(8pi//7) . The sum of these roots is " " sin^(2)""(2pi)/(7) + sin^(2)""(4pi)/(7) + sin ^(2)""(8pi)/(7) = (112)/(64) = (7)/(4) . The value of (tan^(2)""(pi)/(7) + tan^(2)""(2pi)/(7) + tan^(2)""(3pi)/(7))/(cot^(2)""(pi)/(7) + cot^(2)""(2pi)/(7) + cot^(2)""(3pi)/(7)) is

    To find the sum sin^(2) ""(2pi)/(7) + sin^(2)""(4pi)/(7) +sin^(2)""(8pi)/(7) , we follow the following method. Put 7theta = 2npi , where n is any integer. Then " " sin 4 theta = sin( 2npi - 3theta) = - sin 3theta This means that sin theta takes the values 0, pm sin (2pi//7), pmsin(2pi//7), pm sin(4pi//7), and pm sin (8pi//7) . From Eq. (i), we now get " " 2 sin 2 theta cos 2theta = 4 sin^(3) theta - 3 sin theta or 4 sin theta cos theta (1-2 sin^(2) theta)= sin theta ( 4sin ^(2) theta -3) Rejecting the value sin theta =0 , we get " " 4 cos theta (1-2 sin^(2) theta ) = 4 sin ^(2) theta - 3 or 16 cos^(2) theta (1-2 sin^(2) theta)^(2) = ( 4sin ^(2) theta -3)^(2) or 16(1-sin^(2) theta) (1-4 sin^(2) theta + 4 sin ^(4) theta) " " = 16 sin ^(4) theta - 24 sin ^(2) theta +9 or " " 64 sin^(6) theta - 112 sin^(4) theta - 56 sin^(2) theta -7 =0 This is cubic in sin^(2) theta with the roots sin^(2)( 2pi//7), sin^(2) (4pi//7), and sin^(2)(8pi//7) . The sum of these roots is " " sin^(2)""(2pi)/(7) + sin^(2)""(4pi)/(7) + sin ^(2)""(8pi)/(7) = (112)/(64) = (7)/(4) . The value of tan^(2)""(pi)/(7)tan ^(2)""(2pi)/(7) tan ^(2)""(3pi)/(7) is

    Evaluate the following: 1. sin^(-1)(s i n(2pi)/3) 2. cos^(-1)(c o s(7pi)/6) 3. tan^(-1)(t a n(2pi)/3)