Question
In $\triangle\text{ABC}$ prove that, it $\theta$ be any angle, then $\text{b}\cos\theta=\text{c}\cos(\text{A}-\theta)+\text{a}\cos(\text{C}+\theta).$

Answer

$\text{b}\cos\theta=\text{c}\cos(\text{A}-\theta)+\text{a}\cos(\text{C}+\theta)$ Let $\text{a}\sin\text{C = c}\sin\text{A}$ [Using sine rule] $\text{RHS}=\text{c}\cos(\text{A}-\theta)+\text{a}\cos(\text{C}+\theta)$ $=\text{c}\cos\text{A}\cos\theta+\text{c}\sin\text{A}\cos\theta+\text{a}\cos\text{C}.\cos\theta-\text{a}\sin\text{C}\sin\theta$ $=\text{k}\sin\text{C}\cos\text{A}\cos\theta+\text{k}\sin\text{C}\sin\text{A}\cos\theta+\text{k}\sin\text{A}\cos\text{C}.\cos\theta\\-\text{k}\sin\text{A}\sin\text{C}\sin\theta$ $=\text{k}\sin\text{C}\cos\text{A}.\cos\theta+\text{k}\sin\text{A}\cos\text{C}.\cos\theta$ $=\text{k}\cos\theta(\sin\text{C}\cos\text{A}+\sin\text{A}\cos\text{C})$ $=\text{k}\cos\theta\sin(\text{C + A})$ $=\text{k}\cos\theta\sin(\pi-\text{B})$ $=\text{k}\cos\theta\sin\text{B}$ $=\text{k}\sin\text{B}.\cos\theta=\text{b}\cos\theta=\text{LHS}$

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