Trigonometric equations

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$\frac{\sec^{2} θ - \tan θ}{\sec^{2} θ + \tan θ}$ $l i e s b e t w e e n$ $[b, a], t h e n 3 b + a =$

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Solution

$\begin{array}{l} L e t y = \frac{\sec^{2} θ - \tan θ}{\sec^{2} θ + \tan θ} \\ y = \frac{1 + \tan^{2} θ - \tan θ}{1 + \tan^{2} θ + \tan θ} \\ \Rightarrow y + y \tan^{2} θ + y \tan θ = 1 + \tan^{2} θ - \tan θ \\ \Rightarrow (y - 1) \tan^{2} θ + (y + 1) \tan θ + (y - 1) = 0 \end{array}$
Since tan $θ$ is real, we have $b^{2} - 4 a c \geq 0$
$\Rightarrow {(y + 1)}^{2} - 4 {(y - 1)}^{2} \geq 0$
$\Rightarrow 4 {(y - 1)}^{2} - {(y + 1)}^{2} \leq 0$
$\Rightarrow 3 y^{2} - 10 y + 3 \leq 0$
$\Rightarrow \frac{1}{3} \leq y \leq 3$
$H e r e b = \frac{1}{3}, a = 3 ∴ 3 b + a = 4$

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Trigonometric equations

Remember concepts with our Masterclasses.

sec2θ−tanθsec2θ+tanθ lies betweenb,a, then 3b+a=

Let y=sec2θ−tanθsec2θ+tanθ y=1+tan2θ−tanθ1+tan2θ+tanθ⇒y+ytan2θ+ytanθ=1+tan2θ−tanθ⇒(y−1)tan2θ+(y+1)tanθ+(y−1)=0Since tanθ is real, we have b2-4ac≥0⇒(y+1)2−4(y−1)2≥0⇒4(y−1)2−(y+1)2≤0⇒3y2−10y+3≤0⇒13≤y≤3Here b=13,a=3 ∴3b+a=4

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$\frac{\sec^{2} θ - \tan θ}{\sec^{2} θ + \tan θ}$ $l i e s b e t w e e n$ $[b, a], t h e n 3 b + a =$