In a triangle ABC prove that for any angle `theta b cos (A - theta) + a cos (B + theta) = C cos theta`.

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1.	In a triangle ABC, prove that for any angle `theta, b cos (A - theta) + a cos (B + theta) = C cos theta`.
Answer» In `triangleABC` Taking LHS,`bcos(A-theta)+acos(B+theta)` `costheta(bcosAcostheta+sinAsintheta)+a(cosBcostheta-sinBsintheta)` `costheta(bcosA+acosB)+bsinAsintheta-asinBsintheta` `costheta[1/2C(b^2+c^2-a^2+a^2+c^2-b^2)]` `costheta(1/(2C)(2C^2))` We get, `c costheta` which is RHS.

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In a triangle ABC, prove that for any angle `theta, b cos (A - theta) + a cos (B + theta) = C cos theta`.