Given that: $ (1+\cos \alpha)(1+\cos \beta)(1+\cos \gamma)=(1-\cos \alpha)(1-\cos \beta)(1-\cos \gamma) $Show that one of the values of each member of this equality is $ \sin \alpha \sin \beta \sin \gamma $

Given:

\( (1+\cos \alpha)(1+\cos \beta)(1+\cos \gamma)=(1-\cos \alpha)(1-\cos \beta)(1-\cos \gamma) \)

To do:

We have to show that one of the values of each member of the given equality is \( \sin \alpha \sin \beta \sin \gamma \).

Solution:  

$(1+\cos \alpha)(1+\cos \beta)(1+\cos \gamma)=(1-\cos \alpha)(1-\cos \beta)(1-\cos \gamma)$

$\Rightarrow  \frac{(1+\cos \alpha)(1+\cos \beta)(1+\cos \gamma)}{(1-\cos \alpha)(1-\cos \beta)(1-\cos \gamma)}=1$

$\Rightarrow (\frac{1+\cos \alpha}{1-\cos \alpha}) (\frac{1+\cos \beta}{1-\cos \beta}) (\frac{1+\cos \gamma}{1-\cos \gamma})=1$

$\Rightarrow \frac{(1+\cos \alpha)(1+\cos \alpha)}{(1-\cos \alpha)(1+\cos \alpha)} \frac{(1+\cos \beta)(1+\cos \beta)}{(1-\cos \beta)(1+\cos \beta)}\frac{(1+\cos \gamma)(1+\cos \gamma)}{(1-\cos \gamma)(1+\cos \gamma)}=1$

$\Rightarrow  \frac{(1+\cos \alpha)^{2}}{1-\cos ^{2} \alpha}  \frac{(1+\cos \beta)^{2}}{1-\cos ^{2} \beta}  \frac{(1+\cos \gamma)^{2}}{1-\cos ^{2} \gamma}=1$
 $\Rightarrow \frac{(1+\cos \alpha)^{2}}{\sin ^{2} \alpha} \frac{(1+\cos \beta)^{2}}{\sin ^{2} \beta} \frac{(1+\cos \gamma)^{2}}{\sin ^{2} \gamma}=1$
$\Rightarrow (1+\cos \alpha)^{2} (1+\cos \beta)^{2} (1+\cos \gamma)^{2}=\sin ^{2} \alpha \sin ^{2} \beta \sin ^{2} \gamma$
$\Rightarrow (1+\cos \alpha)(1+\cos \beta)(1+\cos \gamma)=\sin \alpha \sin \beta \sin \gamma$

Hence, one of the values of each member of the given equality is \( \sin \alpha \sin \beta \sin \gamma \).