# 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 $\sin \alpha \sin \beta \sin \gamma$.

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