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`Sin^-1x=Pi/6+Cos^-1x` - Mathematics

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प्रश्न

`sin^-1x=pi/6+cos^-1x`

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उत्तर

`sin^-1x=pi/6+cos^-1x`

⇒ `sin^-1x=pi/6+pi/2-sin^-1x`      `[becausecos^-1x=pi/2-sin^-1x]`

⇒ `2sin^-1x=(2pi)/3`

⇒ `sin^-1x=pi/3`

⇒ `sin^-1x=pi/3`

⇒ `x=sin  pi/3=sqrt3/2`

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पाठ 4: Inverse Trigonometric Functions - Exercise 4.10 [पृष्ठ ६६]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 4 Inverse Trigonometric Functions
Exercise 4.10 | Q 7 | पृष्ठ ६६

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