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प्रश्न
Simplify: `cos^-1x + cos^-1 [x/2 + sqrt(3 - 3x^2)/2]; 1/2 ≤ x ≤ 1`
योग
उत्तर
`cos^-1x + cos^-1 [x/2 + sqrt(3 - 3x^2)/2]' 1/2 ≤ x ≤ 1`
Let `cos^-1 x = alpha, x = cos alpha`
Now, = `alpha + cos^-1 [(cos alpha)/2 + (sqrt(3 -3 cos^2 alpha))/2]`
= `alpha + cos^-1 [cos alpha/2 + (sqrt3 .sqrt(1 - cos^2 alpha))/2]`
= `alpha + cos^-1 [cos alpha/2 + sqrt3/2 sin alpha]`
= `alpha + cos^-1 [cos pi/3 cos alpha + sin pi/3 sin alpha]`
= `alpha + cos^-1 [cos(pi/3 - alpha)]`
= `pi/3`
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