Question

If $A,B$ and $C$ are the interior angles of a triangle $ABC$, Prove that:

$\frac{\cos \frac{B+C}{2}\cos \frac{C+A}{2}\cos \frac{A+B}{2}}{\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}}=1$

Answer 1

In $\Delta ABC$,

$\angle A+\angle B+\angle C={180}^o$ ...Angle sum property of triangle

$\Rightarrow \frac{\angle A+\angle B+\angle C}{2}=\frac{{180}^o}{2}$

$\Rightarrow \frac{\angle A+\angle B+\angle C}{2}={90}^o$

$\Rightarrow \frac{\angle A+\angle B}{2}={90}^o-\frac{\angle C}{2}$

Similarly, $\frac{\angle B+\angle C}{2}={90}^o-\frac{\angle A}{2}$ and $\frac{\angle A+\angle C}{2}={90}^o-\frac{\angle B}{2}$

L.H.S $=\frac{\cos \frac{B+C}{2}\cos \frac{C+A}{2}\cos \frac{A+B}{2}}{\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}}$

$=\frac{\cos [{90}^o-\frac{A}{2}]\cos [{90}^o-\frac{B}{2}]\cos [{90}^o-\frac{C}{2}]}{\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}}$

$=\frac{\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}}{\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}}$ ....[$\cos (90-\theta )=\sin \theta$]

$=1$

$=$ R.H.S

Hence proved.