If $$(1+sin\alpha)(1+sin\beta)(1+sin\gamma)=(1-sin\alpha)(1-sin\beta)(1-sin\gamma)$$ then each side is equal to
Let $$(1+sin\alpha)(1+sin\beta)(1+sin\gamma)=k$$ …. (1)
and $$(1-sin\alpha)(1-sin\beta)(1-sin\gamma)=k$$k ….. (2)
now (1)×(2) gives
$$(1^{2}-sinα^{2})(1²-sinβ ²)(1²-sinγ ²)=k²$$
cosα² cosβ² cosγ² = k²
Hence , k= $$\pm cos \alpha cos\beta cos\gamma$$
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