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If 3/(2+costheta+isintheta)=x+iy then (x...

If `3/(2+costheta+isintheta)=x+iy` then (x-1)(x-3)=

A

`y^2`

B

`-y^2`

C

0

D

1

Text Solution

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The correct Answer is:
B
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Knowledge Check

  • |1+costheta+isintheta|=

    A
    `cos(theta//2)`
    B
    `sin(theta//2)`
    C
    `2sin(theta//2)`
    D
    `2cos(theta//2)`

  • (1-costheta+isintheta)^8 =

    A
    `256sin^8(theta//2)[cos4theta+isin4theta]`
    B
    `256cos^8(theta//2)[cos4theta-isin4theta]`
    C
    `256sin^8(theta//2)[sin4theta-icos4theta]`
    D
    `256sin^8(theta//2)[cos4theta-isin4theta]`

  • (1-costheta+isintheta)^6 =

    A
    `2^6sin^6(theta)/(2)(icos3theta+sin3theta)`
    B
    `2^6sin^6(theta)/(2)(sin3theta-icos3theta)`
    C
    `2^6sin^6(theta)/(2)(cot3theta+isin3theta)`
    D
    `2^6sin^6(theta)/(2)(-cot3theta+isin3theta)`

    • Similar Questions

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    If z=costheta+isintheta , then (z^(2n)-1)/(z^(2n)+1)

    If x +iy = (3)/(2 + cos theta + isin theta) , show that x^(2) + y^(2) = 4x - 3

    (1+costheta-isintheta)^4 =

    Match the following. {:(Column-I," "Column-II),(I" : "x=costheta+isintheta" then "x^n-(1)/(x^n),"a) "2isin(alpha-beta)),(II" : If " z=costheta+sintheta" then "(z^n-1)/(z^(2n)+1),"b) "2cos(alpha-beta)),(III" : "x=cisalpha","y=cisbeta" then "x/y+y/x=,"c) "itanntheta),(IV " : " x=cisalpha","y=cisbeta" then "x/y-y/x=,"d) "2iSinntheta):}

    If (1+costheta-isintheta)(1+cos2theta+isin2theta)=x+iy then x^2+y^2 =

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