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प्रश्न
Solve the following equation: `(cos^2θ - 3 cosθ + 2)/sin^2θ` = 1.
उत्तर
We have,
`(cos^2 θ - 3cos θ + 2)/(sin^2 θ) = 1`
⇒ cos2 θ - 3cos θ + 2 = sin2 θ
⇒ cos2 θ - 3cos θ + 2 - sin2 θ = 0
⇒ cos2 θ - 3cos θ + 1 + cos2 θ = 0
⇒ 2cos2 θ - 3cos θ + 1 = 0
⇒ 2cos2 θ - 2cos θ - cos θ + 1 = 0
⇒ 2cos θ( cos θ - 1) - 1( cos θ - 1) = 0
⇒ (cos θ - 1)(2cos θ - 1) = 0
⇒ cos θ - 1 = 0 or ⇒ 2cos θ - 1 = 0
⇒ cos θ = 1 or ⇒ cos θ = `1/2`
⇒ θ = 0° 0r ⇒ θ = 60°
Since, 0 < θ < 90°
So, θ = 60° is the solution of the equation.
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