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If cos–1α + cos–1β + cos–1γ = 3π, then α(β + γ) + β(γ + α) + γ(α + β) equals ______. - Mathematics

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प्रश्न

If cos–1α + cos–1β + cos–1γ = 3π, then α(β + γ) + β(γ + α) + γ(α + β) equals ______.

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उत्तर

If cos–1α + cos–1β + cos–1γ = 3π, then α(β + γ) + β(γ + α) + γ(α + β) equals 6.

Explanation:

We have cos–1α + cos–1β + cos–1γ = 3π

⇒ cos–1α + cos–1β + cos–1γ = π + π + π

⇒ cos–1α = π, cos–1β = π and cos–1γ = π

⇒ α = cos π, β = cos π and γ = cos π

∴ α = – 1, β = – 1 and γ = – 1

Which gives α = β = γ = –1

So α(β + γ) + β(γ + α) + γ(α + β)

⇒ (– 1)(– 1 – 1) + (– 1)(– 1 – 1) + (– 1)(– 1 – 1)

⇒ (– 1)(– 2) + (– 1)(– 2) + (– 1)(– 2)

⇒ 2 + 2 + 2

⇒ 6

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अध्याय 2: Inverse Trigonometric Functions - Exercise [पृष्ठ ३९]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 2 Inverse Trigonometric Functions
Exercise | Q 35 | पृष्ठ ३९

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