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`4sin^-1x=Pi-cos^-1x`

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Question

`4sin^-1x=pi-cos^-1x`

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Solution

`4sin^-1x=pi-cos^-1x`

⇒ `4sin^-1x=pi-(pi/2-sin^-1x)`      `[becausecos^-1x=pi/2-sin^-1x]`

⇒ `4sin^-1x=pi/2+sin^-1x`

⇒ `3sin^-1x=pi/2`

⇒ `sin^-1x=pi/6`

⇒ `x=sin  pi/6=1/2`

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Chapter 3: Inverse Trigonometric Functions - Exercise 4.10 [Page 66]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 3 Inverse Trigonometric Functions
Exercise 4.10 | Q 8 | Page 66
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