Home
Class 11
MATHS
if costheta , sinphi , sintheta are in g...

if cos`theta` , sin`phi` , sin`theta` are in g.p then check the nature of roots of `x^2+2cotphi.x + 1=0`

Promotional Banner

Similar Questions

Explore conceptually related problems

If sintheta,tantheta,costheta are in G.P. then 4sin^2theta-3sin^4theta+sin^6theta=?

Solve : 3-2 cos theta -4 sin theta - cos 2theta+sin 2theta=0 .

Show that ((cos theta - cos 3theta)(sin 8theta + sin 2 theta))/((sin 5 theta - sin theta)(cos 4 theta - cos 6 theta)) = 1

if a=cos theta + i sin theta , b=cos 2 theta- i sin 2 theta , c= cos 3theta + i sin 3theta and if |{:(a,,b,,c),(b,,c,,a),(c,,a,,b):}| =0 then

The expression cos 3 theta + sin 3 theta + (2 sin 2 theta-3) (sin theta- cos theta) is positive for all theta in

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

Consider the cubic equation x^3-(1+cos theta+sin theta)x^2+(cos theta sin theta+cos theta+sin theta)x-sin theta. cos theta =0 Whose roots are x_1, x_2 and x_3

cos 2theta cos 2phi + sin^(2) (theta - phi ) - sin^(2) ( theta + phi) is equal to