Question

Prove that :
${\cos }^{2}A+{\cos }^{2}B-2\cos A\cos B\cos \left(A+B\right)=si{n}^2\left(A+B\right)$

Answer 1

$\begin{array}{c}L.H.S={\cos }^{2}A+{\cos }^{2}B-2\cos A\cos B\cos \left(A+B\right)\\ ={\cos }^{2}A+{\cos }^{2}B-2\cos A\cos B\left(\cos A\cos B-\sin A\sin B\right)\\ ={\cos }^{2}A+{\cos }^{2}B-2{\cos }^{2}A{\cos }^{2}B+2\cos A\sin AcosBsinB\\ R.H.S=si{n}^2\left(A+B\right)\\ ={\left(\sin \left(A+B\right)\right)}^2\\ ={\left(\sin A\cos B+\cos AsinB\right)}^2\\ ={\sin }^{2}A{\cos }^{2}B+{\cos }^{2}A{\sin }^{2}B+2\sin A\cos A\sin B\cos B\\ =\left(1-{\cos }^{2}A\right){\cos }^{2}B+{\cos }^{2}A-{\cos }^{2}A{\cos }^{2}B\\ ={\cos }^{2}B-{\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}A-{\cos }^{2}A{\cos }^{2}B.\\ ={\cos }^{2}A+{\cos }^{2}B-2{\cos }^{2}A{\cos }^{2}B+2\cos A\sin A\cos BSinB\\ Hence,\\ L.H.S=R.H.S\end{array}$