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प्रश्न
Find the value of the expression in terms of x, with the help of a reference triangle
`tan(sin^-1(x + 1/2))`
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उत्तर
Let θ = `tan(sin^-1(x + 1/2))`
⇒ sin θ = `x + 1/2`
1 – sin2θ = `1 - (x + 1/2)^2`
cos2θ = `1 - x^2 + x - 1/4`
cos2θ = `3/4 + x - x^2`
cos2θ = `(3 + 4x - 4x^2)/4`
cos θ = `sqrt(3 + 4x + 4x^2)/2`
tan θ = `((x + 1/2) xx 2)/sqrt(3 + 4x + 4x^2)`
tan θ = `(2x + 1)/sqrt(3 + 4x - 4x^2)`
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