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If tantheta=a/b , then (asintheta+bcosth...

If `tantheta=a/b` , then `(asintheta+bcostheta)/(asintheta-bcostheta)` is equal to (a) `(a^2+b^2)/(a^2-b^2)` (b) `(a^2-b^2)/(a^2+b^2)` (c) `(a+b)/(a-b)` (d) `(a-b)/(a+b)`

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Knowledge Check

  • If btantheta=a , the value of (asintheta-bcostheta)/(asintheta+bcostheta) (a) (a-b)/(a^2+b^2) (b) (a+b)/(a^2+b^2) (c) (a^2+b^2)/(a^2-b^2) (d) (a^2-b^2)/(a^2+b^2)

    A
    `(a-b)/(a^2+b^2)`
    B
    `(a+b)/(a^2+b^2)`
    C
    `(a^2+b^2)/(a^2-b^2)`
    D
    `(a^2-b^2)/(a^2+b^2)`

  • If asintheta+bcostheta=c , then acostheta-b sintheta is equal to

    A
    `+-sqrt(a+b+c)`
    B
    `+-sqrt(a^(2)+b^(2)+c^(2))`
    C
    `+-sqrt(c^(2)+a^(2)-b^(2))`
    D
    `+-sqrt(c^(2)+a^(2)-b^(2))`

  • If x=acostheta+bsinthetaandy=bcostheta-asintheta ,then x^(2)+y^(2) is equal to

    A
    ab
    B
    `a^(2)+b^(2)`
    C
    `a^(2)-b^(2)`
    D
    1

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    If acostheta+bsintheta=c , then prove that: asintheta-bcostheta=+-sqrt(a^(2)+b^(2)-c^(2))

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