If (3-tan^(2)""(pi)/(7))/(1-tan^(2)""(pi...

If `(3-tan^(2)""(pi)/(7))/(1-tan^(2)""(pi)/(7))=kcos""(pi)/(7)` then the value of k is

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The correct Answer is:

Let `theta=(pi)/(7)`
`rArr 3theta=pi-4theta`
`rArr sin3 theta=sin4theta`
`rArr 3sin theta-4 sin^(3)theta=4sinthetacos theta cos 2theta`
`rARr 3-4sin^(2)theta=4cos theta(2cos^(2)theta-1)`
`rArr 8cos^(3)theta-4cos theta=4cos^(2)theta-1`
`rArr 4costheta=(4cos^(2)theta-1)/(2cos^(2)theta-1)=(3-tan^(2)theta)/(1-tan^(2)theta)`

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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))xx (cot^(2)""(pi)/(7) + cot^(2)""(2pi)/(7) + cot^(2)""(3pi)/(7)) is

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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

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