Question

If the expression $\frac{(A+B)(B+C)(C+A)}{ABC}$ is equal to $k\cos \left(\frac{\alpha -\beta }{2}\right)\cos \left(\frac{\beta -\gamma }{2}\right)\cos \left(\frac{\gamma -\alpha }{2}\right),$ then k equals

• A. 4
• B. 8
• C. 16
• D. None of these

Answer 1

Answer: (B)

8

$A=e^{i\alpha },B=e^{i\beta },C=e^{i\gamma },$
$\therefore A+B=e^{i\alpha }+e^{i\beta }=(\cos \alpha +\cos \beta )+i(\sin \alpha +\sin \beta )=2\cos \left(\frac{\alpha -\beta }{2}\right)e^{i\left(\frac{\alpha +\beta }{2}\right)}$
Similarly $B+C=2\cos \left(\frac{\beta -\gamma }{2}\right)e^{i(\frac{\beta +\gamma }{2}})$ & $C+A=2\cos \left(\frac{\gamma -\alpha }{2}\right)e^{i(\frac{\gamma +\alpha }{2}})$
$\therefore \frac{(A+B)(B+C)(C+A)}{ABC}$
$=\frac{8\cos \left(\frac{\alpha -\beta }{2}\right)cos\left(\frac{\beta -\gamma }{2}\right)cos\left(\frac{\alpha -\beta }{2}\right)\cdot e^{i(\alpha +\beta +\gamma )}}{e^{i(\alpha +\beta +\gamma )}}$
$=k\cos \left(\frac{\alpha -\beta }{2}\right)cos\left(\frac{\beta -\gamma }{2}\right)cos\left(\frac{\alpha -\beta }{2}\right)$
$\Rightarrow k=8$