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Prove `(9π)/8 - 9/4 Sin^(-1) 1/3 = 9/4 Sin^(-1) (2sqrt2)/3` - Mathematics

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Question

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

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Solution

L.H.S  = `(9pi)/8 - 9/4 sin^(-1)  1/3`

`= 9/4 (pi/2 - sin^(-1)  1/3)`

`= 9/4 (cos^(-1)  1/3)`     ....(1)  `[sin^(-1)x + cos^(-1) x = pi/2]`

Now, let `cos^(-1)  1/3 = x` Then, `cos x  =  1/3 => sin x  =  sqrt(1 -  (1/3)^2) = (2sqrt2)/3` 

`:. x = sin^(-1) (2sqrt2)/3 =>  cos^(-1)  1/3 = sin^(-1)  (2sqrt2)/3`

:. L.H.S =  `9/4 sin^(-1)  (2(sqrt2))/3` = R.H.S

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Chapter 2: Inverse Trigonometric Functions - Exercise 2.3 [Page 52]

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NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 2 Inverse Trigonometric Functions
Exercise 2.3 | Q 12 | Page 52

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