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प्रश्न
Solve the equation for x:
sin-1x + sin-1(1 - x) = cos-1x, x ≠ 0
बेरीज
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उत्तर
sin-1x + sin-1(1 - x) = cos-1x
`=>"sin"^-1{"x"sqrt(1-(1-"x")^2) + (1-"x")sqrt(1-"x"^2)} = "sin"^-1sqrt(1-"x"^2)`
`[∵ "sin"^-1"x" + "sin"^-1"y" = "sin"^-1{"x"sqrt(1-"y"^2) + "y"sqrt(1-"x"^2)} "and" "cos"^-1"x" = "sin"^-1 sqrt(1-"x"^2)]`
`=> "x"sqrt(1-1+2"x" - "x"^2) + sqrt(1-"x"^2) - "x"sqrt(1-"x"^2) = sqrt(1-"x"^2)`
`=> "x"sqrt(2"x"-"x"^2) - "x" sqrt(1-"x"^2) = 0`
`=> "x"(sqrt(2"x"-"x"^2) - sqrt(1-"x"^2))=0`
`=>"x" = 0 , sqrt(2"x" - "x"^2) - sqrt(1-"x"^2) = 0`
`=> x=0 , sqrt(2"x" - "x"^2) = sqrt(1-"x"^2)`
Now, `sqrt(2"x" - "x"^2) = sqrt(1-"x"^2)`
Squaring both sides, we obtain
2x - x2 = 1 - x2
⇒ 2x = 1
⇒ x = `1/2`
Hence, x = 0 and x = `1/2`
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