Question

Given that: (1 + cosα) (1 + Cosβ) (1 + cosγ) = (1 - cosα)(1 – cosβ) (1 – cos γ) .Show that one of the values of each member of this equality is sinα sinβ sinγ.

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Answer 1

Step-by-step explanation:

We have,

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

Multiply both by $(1-\cos(\alpha))(1-\cos(\beta))(1-\cos(\gamma))$

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

$\implies \: (1 - \cos^{2} ( \alpha ))(1 - \cos ^{2} ( \beta ))(1 - \cos^{2} ( \gamma)) = (1 - \cos( \alpha ) )^{2} (1 - \cos( \beta ))^{2} (1 - \cos( \gamma)) ^{2}$

$\implies \: \sin^{2} ( \alpha )\sin^{2} ( \beta ) \sin^{2} ( \gamma) = (1 - \cos( \alpha ) )^{2} (1 - \cos( \beta ))^{2} (1 - \cos( \gamma)) ^{2}$

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