The range of 11−tan2x is

Put x=0°Minimum value =11−0=1x=45°Maximum value =11−1=10=∞[0,∞).

The range of 11−tan2x is

Put x=0°Minimum value =11−0=1x=45°Maximum value =11−1=10=∞[0,∞).

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The range of $\frac{1}{\sqrt{1 - \tan^{2} x}}$ is

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$(1, \infty)$

$[1, \infty)$

$[0, \infty)$

$(0, \infty)$

answer is D.

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Detailed Solution

Put $x = 0 °$

Minimum value $= \frac{1}{\sqrt{1 - 0}} = 1$

$x = 45 °$

Maximum value $= \frac{1}{\sqrt{1 - 1}} = \frac{1}{0} = \infty$

$[0, \infty)$ .

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