Practicing Success
If $cos^2θ -sin^2θ - 3 cosθ + 2, 0° < θ < 90°, $ then what will be the value of secθ - cosθ ? |
$\frac{4}{3}$ $\frac{1}{2}$ $\frac{3}{2}$ $\frac{2}{3}$ |
$\frac{3}{2}$ |
cos²θ - sin²θ - 3cosθ + 2 = 0 cos²θ - ( 1 - cos²θ) - 3cosθ + 2 = 0 2cos²θ - 3cosθ + 1 = 0 2cos²θ - 2cosθ - cosθ + 1 = 0 2cosθ ( cosθ - 1 ) - 1 ( cosθ - 1 ) = 0 ( 2cosθ - 1 ) (cosθ - 1 ) = 0 Either 2cosθ - 1 = 0 OR cosθ - 1 = 0 But θ < 90º So, 2cosθ - 1 = 0 cosθ = \(\frac{1}{2}\) { cos60º = \(\frac{1}{2}\) } So, θ = 60º Now, secθ - cosθ = sec60º - cos60º = 2 - \(\frac{1}{2}\) = \(\frac{3}{2}\)
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