Question

If : (1 + cos α) × (1 + Cosβ) × (1 + cosγ) = (1 - cosα) × (1 – cosβ) × (1 – cos γ) . Then, show that one of the values of each member of the given equality is sinα sinβ and sinγ.​

Answer 1

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$\implies$$\large\bf{\underline{\red{VERIFIED✔}}}$

$We have:  (1+cosα)(1+cosβ)(1+cosγ) \\ (1−cosα)(1−cosβ)(1−cosγ) \\ Multiplying  both sides by \\ (1+cosα)(1+cosβ)(1+cosγ), we get \\ (1+cosα)²(1+cosβ)²(1+cosγ)² \\ (1−cosα)(1−cosβ)(1−cosγ)(1+cosα) \\ (1+cosβ)(1+cosγ) \\ ⇒(1+cosα)²(1+cosβ)²(1+cosγ)²\\ =(1−cos²α)(1−cos²β)(1−cos²γ) \\⇒(1+cosα)²(1+cosβ)²(1+cosγ)² \\=sin²α sin²β sin²γ \\ ⇒(1+cosα)(1+cosβ)(1+cosγ)=±sin\:α\:sin\:β\:sin\:γ \\ Hence, \:one \:of\: the\: values \:of  \\ (1+cosα)(1+cosβ)(1+cosγ) is sin \: α \: sin \: β \: sin \: γ$

$\boxed{I \:Hope\: it's \:Helpful}$

${\sf{\bf{\blue{@ℐᴛz ᴅɪɴᴜ࿐}}}}$

Answer 2

$Wehave:  (1+cosα)(1+cosβ)(1+cosγ)(1−cosα)(1−cosβ)(1−cosγ)Multiplying bothsidesby(1+cosα)(1+cosβ)(1+cosγ),weget(1+cosα)²(1+cosβ)²(1+cosγ)²(1−cosα)(1−cosβ)(1−cosγ)(1+cosα)(1+cosβ)(1+cosγ) ⇒(1+cosα)²(1+cosβ)²(1+cosγ)²=(1−cos²α)(1−cos²β)(1−cos²γ)⇒(1+cosα)²(1+cosβ)²(1+cosγ)²=sin²αsin²βsin²γ⇒(1+cosα)(1+cosβ)(1+cosγ)=±sinαsinβsinγHence,oneofthevaluesof (1+cosα)(1+cosβ)(1+cosγ) is sinαsinβsinγ</p><p></p><p> \boxed{I \:Hope\: it's \:Helpful} </p><p>$