Question

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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

Required Answer:

Question:

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

Solution:

We have;

➡️ (1 + cosα) (1 + Cosβ) (1 + cosγ) = (1 - cosα)(1 – cosβ) (1 – cosγ)

Multiplying both sides by (1+Cosa)(1+cosβ)(1+cosγ), we get;

➡️[(1+cosa)(1+cosβ)(1+cosγ)]^2= (1 - cosα)(1 – cosβ)(1 – cosγ) × (1+cosa)(1+cosβ)(1+cosγ)

➡️[(1+cosa)(1+cosβ)(1+cosγ)]^2 =

(1-cos^2a)(1- cos^2β)(1- cos^2γ)

➡️[(1+cosa)(1+cosβ)(1+cosγ)]^2 = (sin^2a)(sin^2β)(sin^2γ)

➡️[(1+cosa)(1+cosβ)(1+cosγ)]^2 = ± sina sinβ sinγ

☆Hence one of the values if (1+ cosa)(1+cosβ)(1+cosγ) is sina sinβ sinγ.

Now again we have,

➡️1 + cosα) (1 + Cosβ) (1 + cosγ) = (1 - cosα)(1 – cosβ) (1 – cos γ)

Multiplying both sides by (1 - cosα)(1 – cosβ) (1 – cosγ), we get;

➡️(1+cosa)(1+cosβ)(1+cosγ)(1 - cosα)(1 – cosβ)

(1 – cosγ)=[ (1 - cosα)(1 – cosβ) (1 – cosγ)]^2

➡️[(1 - cosα)(1 – cosβ) (1 – cosγ)]^2 = (1 - cos^2α)

(1 – cos^2β)(1 – cos^2γ)

➡️[(1 - cosα)(1 – cosβ) (1 – cosγ)]^2 = (sin^2a)(sin^2β)(sin^2γ)

➡️ (1 - cosα)(1 – cosβ) (1 – cosγ) = ± (sina)(sinβ)

(sinγ)

☆Hence one of the values of (1 - cosα)(1 – cosβ) (1 – cosγ) is (sina)(sinβ)(sinγ)

Thus, one of the values of each member of this equality is (sina)(sinβ)(sinγ).

(Hence Proved)

Formula used:-

• sin^2(x) = 1 - cos^2(x).

Answer 2

Answer:

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