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प्रश्न
Solve the following equations:
cot θ + cosec θ = `sqrt(3)`
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उत्तर
cot θ + cosec θ = `sqrt(3)`
`cos theta/sin theta + 1/sin theta = sqrt(3)`, sin θ ≠ 0
`(cos theta + 1)/sin theta = sqrt(3)`, sin θ ≠ 0
1 + cos θ = `sqrt(3) sin theta`
`sqrt(3)sin theta - cos theta` = 1
Divide each term by 2
`sqrt(3)/2 sin theta - 1/2 cos theta = 1/2`
`sin pi/3 * sin theta - cos pi/3 * cos theta = 1/2`
`- (cos theta cos pi/3 - sin theta sin pi/3) = 1/2`
`cos (theta + pi/3) = - 1/2`
`cos (theta + pi/3) = cos (theta - pi/3)`
`cos (theta + pi/3) = cos ((3pi - pi)/3)`
`cos (theta + pi/3) = cos ((2pi)/3)`
The general solution is
`theta + pi/3 = 2"n"pi + (2pi)/3`, n ∈ Z
θ = `2"n"pi - pi/3 + (2pi)/3`, n ∈ Z
θ = `2"n"pi - pi/3 - (2pi)/3` or θ = `2"n"pi - pi/3 + (2pi)/3`
θ = `2"n"pi - (3pi)/3` or θ = `2"n"pi + (2pi - pi)/3`
θ = `2"n"pi - pi` or θ = `2"n"pi + pi/3`, n ∈ Z
θ = `(2"n" - 1)pi` or θ = `2"n"pi + pi/3`, n ∈ Z
Since sin θ ≠ 0, θ = (2n – 1)π is not possible
∴ θ = `2"n"pi + pi/3`, n ∈ Z
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