Question

2) . 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

$We have:  (1+cosα)(1+cosβ)(1+cosγ)</p><p></p><p>(1−cosα)(1−cosβ)(1−cosγ)</p><p></p><p>Multiplying  both sides by</p><p></p><p>(1+cosα)(1+cosβ)(1+cosγ), we get</p><p></p><p>(1+cosα)2(1+cosβ)2(1+cosγ)2</p><p></p><p>(1−cosα)(1−cosβ)(1−cosγ)(1+cosα)(1+cosβ)(1+cosγ) </p><p></p><p>⇒(1+cosα)2(1+cosβ)2(1+cosγ)2</p><p></p><p>=(1−cos2α)(1−cos2β)(1−cos2γ)</p><p></p><p>⇒(1+cosα)2(1+cosβ)2(1+cosγ)2=sin2αsin2βsin2γ</p><p></p><p>⇒(1+cosα)(1+cosβ)(1+cosγ)=±sinαsinβsinγ</p><p></p><p>Hence, one of the values of (1+cosα)(1+cosβ)(1+cosγ) is sinαsinβsinγ</p><p></p><p>$

Answer 2

Wehave: (1+cosα)(1+cosβ)(1+cosγ)(1−cosα)(1−cosβ)(1−cosγ)Multiplying bothsidesby(1+cosα)(1+cosβ)(1+cosγ),weget(1+cosα)2(1+cosβ)2(1+cosγ)2(1−cosα)(1−cosβ)(1−cosγ)(1+cosα)(1+cosβ)(1+cosγ) ⇒(1+cosα)2(1+cosβ)2(1+cosγ)2=(1−cos2α)(1−cos2β)(1−cos2γ)⇒(1+cosα)2(1+cosβ)2(1+cosγ)2=sin2αsin2βsin2γ⇒(1+cosα)(1+cosβ)(1+cosγ)=±sinαsinβsinγHence,oneofthevaluesof (1+cosα)(1+cosβ)(1+cosγ) is sinαsinβsinγ