Question

In $△ABC$ prove that if $\theta$ be any angle, then
$b\cos \theta =c\cos (A-\theta )+a\cos (C+\theta )$.

Answer 1

$R.H.S=k\sin CC\cos (A-\theta )+k\sin A\cos (C+\theta )$
$=k[\sin C\cos A\cos \theta +\sin C\sin A\sin \theta +\sin A\cos C\cos \theta -\sin A\sin C\sin \theta ]$
$=k\cos \theta (\sin A\cos C+\cos A\sin C)$
$=k\cos \theta \sin (A+C)=k\sin B\cos \theta =b\cos \theta$. Hence proved