tan−1⁡(tan⁡1−θ)=1−θ when

$\tan^{- 1} (\tan \sqrt{1 - θ}) = \sqrt{1 - θ}$ when

A
$- \frac{π}{2} < θ < \frac{π}{2}$
B
$θ > \frac{4 - π}{4}$
C
$θ < \frac{4 - π}{4}$
D
$\frac{4 - π^{2}}{4} < θ \leq 1$

Solution:

Clearly, $\sqrt{1 - θ}$ is real, If $θ \leq 1 .$

Now,

$\tan^{- 1} (\tan \sqrt{1 - θ}) = \sqrt{1 - θ}$

$\Rightarrow 0 \leq \sqrt{1 - θ} < \frac{π}{2}$ and $θ \leq 1$

$\Rightarrow 0 \leq 1 - θ < \frac{π^{2}}{4}$ and $θ \leq 1$

$\Rightarrow - 1 \leq - θ \leq \frac{π^{2}}{4} - 1$ and $θ \leq 1$

$\Rightarrow 1 - \frac{π^{2}}{4} < θ \leq 1$ and $θ \leq 1 \Rightarrow \frac{4 - π^{2}}{4} < θ \leq 1$

Related content

CBSE Class 10 Grading System and Marking Scheme 2024

Vocabulary Words For IELTS List, Preparation Tips

Science GK Quiz in Hindi : जानिए विज्ञान से जुड़े महत्वपूर्ण सामान्य ज्ञान

UP Board Class 10 Result 2024 Out Check Your Result

How to Study for NEET from Class 8?

Do or Die Chapters For NEET 2024

List of Phobias: Common Phobias From A to Z

JEE Main 2024 April 6 Question Paper and Answer Key PDF with Solutions- Shift 1, 2

NEET MDS 2024 Result by April 18, Check Expected Cutoff

Letter to Your Friend Describing Your Ancestral House