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.


Discussion

No Comment Found