cot(x-y)= cotxcoty+1/coty-cotx?

I need help with solving this!! I have to verify the identity. Thank you!

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  • 1 decade ago
    Favorite Answer

    LHS

    = cot(x-y)

    = 1/tan(x-y)

    = 1/((tan(x) - tan(y))/(1 + tan(x)tan(y)))

    = (1 + tan(x)tan(y)) / (tan(x) - tan(y)) ------- now divide the numerator and denominator by tan(x)tan(y)

    = (cot(x)cot(y) + 1) / (cot(y) - cot(x))

    = RHS

  • 4 years ago

    1 Cot X

  • 5 years ago

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    RE:

    cot(x-y)= cotxcoty+1/coty-cotx?

    I need help with solving this!! I have to verify the identity. Thank you!

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  • 1 decade ago

    cot(x-y)= cos(x-y)/sin(x-y)=cosx.cosy+sinx.siny/sinx.cosy-cosx.siny

    {divide Nr&Dr by sinx.siny}

    =cotxcoty+1/coty-cotx.=RHS

    Hence proved.

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  • Anonymous
    4 years ago

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    cot(x+y) = (cotxcoty-1)/(cotx+coty)? Girl, please read patiently. It is NOT tough but needs patience. LHS = cot(x+y) = cos(x+y) / sin(x+y) = [(cosx*cosy - sinx*siny) / (sinx*cosy + cosx*siny)] Divide numerator and denominator by sinx*siny, we get (cotxcoty-1)/(coty+cotx), = (cotxcoty-1)/(cotx+coty) = RHS

  • 1 decade ago

    cot(x-y)= cos(x-y)/sin(x-y)

    or = [cosxcosy +sinxsiny]/[sinxcosy-cosxsiny]

    = cotxcoty+1/coty - cotx

    [dividing numeretor and denomerator by sinxsiny]

    now further clarification to prove sin(x-y) and cos(x+y) etc

    one example shown below .others can be found in your text book

    To prove cos(x ± y) = cosx cosy ∓ sinx siny

    then

    sin(x + y) = cos(π/2 - (x+y)) = cos((π/2 - x) - y)

    = cos(π/2 - x) cosy + sin(π/2 - x) siny

    = sinx cosy + cosx siny

    now, to prove that cos(x + y) = cosx cosy - sinx siny .. . .. .

    this used the distance formula:

    let P(1,0), A(cosx, sinx) , B(cosy, siny) , C(cos(x-y), sin(x-y))

    the angle from the positive x-axis to A is x.

    the angle from the positive y-axis to B is y.

    thus the angle between them is x - y , say, x > y.

    thus the distance bet AB is the same distance as CP

    then

    d(AB) = √[(cosx - cosy)^2 + (sinx - siny)^2]

    = √[cos^2x - 2cosx cosy + cos^2y + sin^2x - 2sinxsiny + sin^2y]

    = √[2 - 2(cosx cosy + sinx siny)]

    meanwhile

    d(CP) = √[(1- cos(x-y)^2 + sin^2(x-y)]

    = √[1 - 2cos(x-y) + cos^2(x-y) + sin^2(x-y)]

    = √[2 - 2cos(x-y)]

    thus

    cos(x - y) = (cosx cosy + sinx siny)

  • 1 decade ago

    LHS

    = cot(x-y)

    = 1/tan(x-y)

    = 1/((tan(x) - tan(y))/(1 + tan(x)tan(y))) {(formula of tan(x-y)}

    = (1 + tan(x)tan(y)) / (tan(x) - tan(y)) ------- now divide the numerator and denominator by tan(x)tan(y)

    = (cot(x)cot(y) + 1) / (cot(y) - cot(x))

    = RHS

    It can use as formula also

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