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Question

In ∆ABC, prove that if θ be any angle, then b cosθ = c cos (A − θ) + a cos (C + θ).

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Solution

Suppose asinA=bsinB=csinC=k. ...(1)

Consider the RHS of the equation b cosθ = c cos (A − θ) + a cos (C + θ).

RHS=ccosA-θ+acosC+θ =ksinCcosA-θ+ksinAcosC+θ from 1 =k22sinCcosA-θ+2sinAcosC+θ =k2sinA+C-θ+sinC+θ-A+sinA+C+θ+sinA-C-θ =k2sinπ-B-θ+sinC+θ-A+sinπ-B+θ-sinC+θ-A A+B+C=π =k2sinB+θ+sinB-θ =k2sinBcosθ+sinθcosB+sinBcosθ-sinθcosB =k22sinBcosθ =ksinBcosθ =bcosθ=LHSHence proved.

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