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

Required Answer:-

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.

Solution:

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)(1 + Cosβ)(1 + cosγ)]² = (1 - cosα)(1 – cosβ)(1 – cosγ) × (1 + cosα)(1 + Cosβ)(1 + cosγ)

➡ [(1 + cosα)(1 + Cosβ)(1 + cosγ)]² = (1 - cos²α)(1 – cos²β)(1 – cos²γ)

➡ [(1 + cosα)(1 + Cosβ)(1 + cosγ)]² = (sin²α)(sin²β)(sin²γ)

➡ [(1 + cosα)(1 + Cosβ)(1 + cosγ)] = ± sinα sinβ sinγ

➡ Hence, one of the values of (1 + cosα) (1 + Cosβ) (1 + cosγ) is sinα sinβ sinγ

Again, 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)(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 - cos²α)(1 – cos²β)(1 – cos²γ)

➡ [(1 - cosα)(1 – cosβ)(1 – cosγ)]² = (sin²α)(sin²β)(sin²γ)

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

➡ Hence, one of the values of (1 - cosα)(1 – cosβ)(1 – cosγ) is (sinα)(sinβ)(sinγ)

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

(Hence Proved)

Formula Used:

• sin²(x) = 1 - cos²(x)

Answer 2

Answer:

Given :-

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

To Find :-

The values of each member of this equality is sinα sinβ sinγ.

SoluTion :-

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

Here,

We will first multiply both sides by (1 + cosα) (1 + Cosβ) (1 + cosγ)

$\tt \: (1 + \cos( \alpha ) ){}^{2} (1 + \cos( \beta )) {}^{2} (1 + \cos( \gamma ) ) {}^{2}$

$\tt \: (1 - { \cos }^{} \alpha )(1 - { \cos }^{} \beta )(1 - \cos \gamma )(1 + \cos( \alpha ) )(1 + \cos( \beta ))(1 + \cos( \gamma ) )$

$\tt \: (1+ \cos( \alpha )) {}^{2} (1+ \cos( \beta ) ) {}^{2} (1+ \cos( \gamma ) ) {}^{2}$

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

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

$\tt \: (1 + \cos( \alpha ) )(1 + \cos( \beta ) )(1 + \cos( \gamma ) ) ± \: \sin \alpha \sin \beta \sin \gamma$

Therefore :-

One of the values of each member of this equality is sinα sinβ sinγ.