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प्रश्न
if `sin(sin^(-1) 1/5 + cos^(-1) x) = 1` then find the value of x
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उत्तर
`sin(sin^(-1) 1/5 + cos^(-1) x ) = 1`
`=> sin (sin^(-1) 1/5) cos(cos^(-1)x) + cos(sin^(-1) 1/5) sin(cos^(-1) x) = 1`
`[sin(A+B) = sin A cosB + cosA sin B]`
`=> 1/5 xx x + cos(sin^-1 1/5) sin(cos^(-1) x) = 1 `
`=> x/5 + cos(sin^(-1) 1/5) sin (cos^(-1) x) = 1` (1)
Now let `sin^(-1) 1/5 = y`
Then, `siny = 1/5 => cos y = sqrt(1 - (1/5)^2) = (2sqrt6)/5 => y = cos^(-1) ((2sqrt6)/5)`
`:. sin^(-1) 1/5 = cos^(-1) ((2sqrt6)/5) ` ...(2)
Let `cos^(-1) x = z`
Then `cos z = x => sin z = sqrt(1-x^2) => z = sin^(-1) (sqrt(1-x^2))`
`:. cos^(-1) x = sin^(-1) (sqrt(1-x^2))`
From 1, 2 and 3 we have
`x/5 + cos(cos^(-1) (2sqrt6)/5). sin(sin^(-1)sqrt(1- x^2)) = 1`
`=> x/5 + (2sqrt6)/5. sqrt(1 - x^2) = 1`
`=> x + 2sqrt6sqrt(1-x^2) = 5`
`= 2sqrt6sqrt(1-x^2) = 5 - x`
On squaring both sides, we get:
`(4)(6)(1-x^2) = 25 + x^2 - 10x`
`=> 24 - 24x^2 = 25 + x^2 - 10x`
`=> 25x^2 - 10x + 1 = 0`
`=> (5x - 1)^2 = 0`
=> (5x -1) = 0
`=> x = 1/5`
Hence, the value of x is `1/5`
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