Practicing Success
If $11 \sin^{2} \theta - \cos^{2} \theta + 4 \sin \theta - 4 = 0, 0^\circ < \theta < 90^\circ$, then what is the value of $\frac{\cos 2\theta + \cot 2 \theta}{\sec 2 \theta - \tan 2 \theta}$ |
$\frac{12+7\sqrt{3}}{6}$ $\frac{12+5\sqrt{3}}{3}$ $\frac{10+5\sqrt{3}}{3}$ $\frac{10+7\sqrt{3}}{6}$ |
$\frac{12+7\sqrt{3}}{6}$ |
11 sin²θ - cos²θ + 4 sinθ - 4 = 0 { sin²θ + cos²θ = 1 } 11 sin²θ - (1 - sin²θ) + 4 sinθ - 4 = 0 11 sin²θ - 1 + sin²θ + 4 sinθ - 4 = 0 12 sin²θ + 4 sinθ - 5 = 0 12 sin²θ + 10 sinθ - 6sinθ - 5 = 0 2sinθ ( 6sinθ +5 ) - 1 ( 6sinθ +5 ) = 0 ( 2sinθ - 1 ). ( 6sinθ +5 ) = 0 Either ( 2sinθ - 1 ) = 0 Or ( 6sinθ +5 ) = 0 ( 6sinθ +5 ) = 0 is not possible. So, 2sinθ - 1 = 0 sinθ = \(\frac{1}{2}\) { sin30º = \(\frac{1}{2}\) } So, θ = 30º Now, \(\frac{sin2θ + cot2θ }{sec2θ - tan2θ }\) = \(\frac{sin60º + cot60º}{sec60º - tan60º }\) = \(\frac{1/2 + √3}{2 -√3 }\) = \(\frac{2 + √3}{2√3(2 -√3) }\) = \(\frac{2 + √3}{2√3(2 -√3) }\) × \(\frac{2 + √3}{2 +√3}\) = \(\frac{12 + 7√3}{6 }\)
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