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प्रश्न
Solve the equation for x: `sin^(-1) 5/x + sin^(-1) 12/x = π/2, x ≠ 0`
Solve the equation x: `sin^(-1) (5/x) + sin^(-1) (12/x) = π/2 (x ≠ 0)`
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उत्तर १
`sin^(-1) (5/x) + sin^(-1) (12/x) = π/2`
`sin^(-1) + cos^(-1) sqrt(1 - 144/x^2) = π/2`
Let `sin^(-1) 12/x = β`
`12/x = sin β = "OPP"/"HYP"`
`sqrt(x^2 - 144)/x = cos β = "Adj"/"HYP"`
`β = cos^(-1) (sqrt(x^2 - 144)/x^2)`
∴ `5/x = sqrt(1 - 144/x^2)`
`25/x^2 = 1 - 144/x^2`
`169/(x^2) = 1`
x2 = 169
x = 13
उत्तर २
`sin^(-1) (5/x) + sin^(-1) (12/x) = π/2`
⇒ `sin^-1 (5/x) = π/2 - sin^-1 (12/x)`
⇒ `sin^-1 (5/x) = cos^-1 (12/x)` ...`[∵ sin^-1x + cos^-1x = π/2]`
⇒ `sin^-1 (5/x) = cos^-1 (12/x) = θ`
⇒ `sin θ = 5/x, cos θ = 12/x`
Since, sin2θ + cos2θ = 1
⇒ `(5/x)^2 + (12/x)^2 = 1`
⇒ `25/(x^2) + 144/(x^2) = 1`
⇒ `169/(x^2) = 1`
⇒ x = ±13
Since, x = –13 does not satisfy the given equation.
So, x = 13.
Notes
Students should refer to the answer according to their questions.
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