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If $y = \frac{\sec^{2} θ - \tan θ}{\sec^{2} θ + \tan θ}$ , then

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y∉[1/3,3]

−3

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Correct option is A

We have,sec2⁡θ−tan⁡θsec2⁡θ+tan⁡θ=y⇒ 1+x2−x1+x2+x=y where tan⁡θ=x⇒ x2(y−1)+x(y+1)+y−1=0⇒ (y+1)2−4(y−1)2≥0 ∵x=tan⁡θ is real ∴ Disc ≥0⇒ −3y2+10y−3≥0⇒ 3y2−10y+3≤0⇒y∈(1/3,3)

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If y=sec2⁡θ−tan⁡θsec2⁡θ+tan⁡θ, then

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If $y = \frac{\sec^{2} θ - \tan θ}{\sec^{2} θ + \tan θ}$ , then