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`(9pi)/8-9/4sin^-1 1/3=9/4sin^-1 (2sqrt2)/3` - Mathematics

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Question

`(9pi)/8-9/4sin^-1  1/3=9/4sin^-1  (2sqrt2)/3`

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Solution

`(9pi)/8-9/4sin^-1  1/3=9/4sin^-1  (2sqrt2)/3`

LHS = `(9pi)/8-9/4sin^-1  1/3`

`=9/4(pi/2-sin^-1  1/3)`

`=9/4(cos^-1  1/3)`

`=9/4(sin^-1sqrt(1-1/9))`

`=9/4(sin^-1  (2sqrt2)/3)=` RHS


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Chapter 4: Inverse Trigonometric Functions - Exercise 4.12 [Page 89]

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RD Sharma Mathematics [English] Class 12
Chapter 4 Inverse Trigonometric Functions
Exercise 4.12 | Q 2.3 | Page 89

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