First slide

Extreme values and Periodicity of Trigonometric functions

Question

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

Moderate

A

$\frac{1}{3} < y < 3$
B

$y \notin [1 / 3, 3]$
C

$- 3 < y < - \frac{1}{3}$
D

none of these

Solution

We have,
$\frac{\sec^{2} θ - \tan θ}{\sec^{2} θ + \tan θ} = y$
$\Rightarrow \frac{1 + x^{2} - x}{1 + x^{2} + x} = y$ where $\tan θ = x$
$\begin{array}{l} \Rightarrow x^{2} (y - 1) + x (y + 1) + y - 1 = 0 \\ \Rightarrow (y + 1)^{2} - 4 (y - 1)^{2} \geq 0 [\begin{array}{l} ∵ x = \tan θ is real \\ ∴ Disc \geq 0 \end{array}] \\ \Rightarrow - 3 y^{2} + 10 y - 3 \geq 0 \\ \Rightarrow 3 y^{2} - 10 y + 3 \leq 0 \Rightarrow y \in (1 / 3, 3) \end{array}$

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