Given that:


Given that:

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

Show that one of the values of each member of this equality is $\sin \alpha \sin \beta \sin \gamma$


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

Let us assume that

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

We know that, $\sin ^{2} \theta+\cos ^{2} \theta=1$

Then, we have

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

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

$\Rightarrow L^{2}=\left(1-\cos ^{2} \alpha\right)\left(1-\cos ^{2} \beta\right)\left(1-\cos ^{2} \gamma\right)$

$\Rightarrow L^{2}=\sin ^{2} \alpha \sin ^{2} \beta \sin ^{2} \gamma$


$\Rightarrow L=\pm \sin \alpha \sin \beta \sin \gamma$

Therefore, we have

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

Taking the expression with the positive sign, we have

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

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