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.If α,β are two values of θ satisfying equation cosθ/a+sinθ/b=1/c then.Prove that cot(α+β/2)=b/a.PLease kindly explain with full steps

Akshay
185 Points
5 years ago
cot(α+β/2) = (1+cos(α+β)) / (sin(α+β)) = [1+cosα*cosβ – sinα*sinβ] / [sinα*cosβ + cosα*sinβ] = [1+cosα*cosβ*(1 – tanα*tanβ] / [cosα*cosβ*(tanα + tanβ)] …...eq(1),
take ur eq. cosθ/a+sinθ/b=1/c,
take cosθ/a to right side and then square and substitute sin2θ = 1 – cos
2θ,
you will get quardratic equation in cosθ, find cosα*cos
β using that,
similarly find a quardratic equation in tanθ, for that divide original equation by cosθ, then sqaure and substitute sec2θ = 1+tan2θ on right hand side. using this quardratic equation you can find tanα*tanβ
substitute in equation (1).
pa1
357 Points
5 years ago

cot(α+β/2) = (1+cos(α+β)) / (sin(α+β)) = [1+cosα*cosβ – sinα*sinβ] / [sinα*cosβ + cosα*sinβ] = [1+cosα*cosβ*(1 – tanα*tanβ] / [cosα*cosβ*(tanα + tanβ)] …...eq(1),
take ur eq. cosθ/a+sinθ/b=1/c,
take cosθ/a to right side and then square and substitute sin2θ = 1 – cos
2θ,
you will get quardratic equation in cosθ, find cosα*cos
β using that,
similarly find a quardratic equation in tanθ, for that divide original equation by cosθ, then sqaure and substitute sec2θ = 1+tan2θ on right hand side. using this quardratic equation you can find tanα*tanβ
substitute in equation (1).