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If cos-1 x + cos-1y + cos-1z = 3π, then show that x2 + y2 + z2 + 2xyz = 1.

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प्रश्न

If cos-1 x + cos-1y + cos-1z = 3π, then show that x2 + y2 + z2 + 2xyz = 1.

योग
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उत्तर

0 ≤ cos-1x ≤ π and 

cos-1x + cos-1 y + cos-1z = 3π

∴ cos-1x = π, cos-1y = π  and cos-1z = π

∴ x = y = z = cos π = - 1

∴ x2 + y2 + z2 + 2xyz

= (- 1)2 + (- 1)2 + (- 1)2 + 2(- 1)(- 1)(- 1)

= 1 + 1 + 1 - 2

= 3 - 2

= 1.

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अध्याय 3: Trigonometric Functions - Miscellaneous exercise 3 [पृष्ठ १११]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 3 Trigonometric Functions
Miscellaneous exercise 3 | Q 39 | पृष्ठ १११

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