Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

Given $\sin A\sin B\sin C=\frac{1}{8}$ and $\cos A\cos B\cos C=\frac{1}{8}$. Since $A,B,C$ are acute angles, we can divide the two given equations:$$\frac{\sin A\sin B\sin C}{\cos A\cos B\cos C}=\frac{1/8}{1/8}=1$$ $$\tan A\tan B\tan C=1$$ For acute angles$A,B,C$, the condition $\tan A\tan B\tan C=1$ implies $A+B+C=\frac{\pi }{2}$. Now we need to find the value of $\sin (A+B)\sin (B+C)\sin (C+A)$. Since$A+B+C=\frac{\pi }{2}$:$A+B=\frac{\pi }{2}-C⟹\sin (A+B)=\sin \left(\frac{\pi }{2}-C\right)=\cos C$.$B+C=\frac{\pi }{2}-A⟹\sin (B+C)=\sin \left(\frac{\pi }{2}-A\right)=\cos A$.$C+A=\frac{\pi }{2}-B⟹\sin (C+A)=\sin \left(\frac{\pi }{2}-B\right)=\cos B$. Therefore, $\sin (A+B)\sin (B+C)\sin (C+A)=\cos C\cos A\cos B$. From the given information, $\cos A\cos B\cos C=\frac{1}{8}$. So, the value of the expression is$\frac{1}{8}$.