Question

If $A=B={45}^0$, then

$\cos (A+B)=\cos A\cos B-\sin A\sin B$

If true enter 1 else 0

Answer 1

$\cos (A+B)=\cos ({45}^0+{45}^0)=\cos \left({90}^0\right)=0$ (L.H.S)
$\cos A\cos B-\sin A\sin B$
= $\cos \left({45}^0\right)\cos \left({45}^0\right)-\sin \left({45}^0\right)\sin \left({45}^0\right)$
= $\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)-\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)$
= $\frac{1}{2}-\frac{1}{2}$
= $0$ (R.H.S)

Hence, L.H.S=R.H.S

Hence correct answer 1.