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Question: If \(\theta = \sin^{- 1}x + \cos^{- 1}x - \tan^{- 1}x,x \geq 0\) then the smallest interval in whic...

If θ=sin⁡−1x+cos⁡−1x−tan⁡−1x,x≥0\theta = \sin^{- 1}x + \cos^{- 1}x - \tan^{- 1}x,x \geq 0 then the smallest

interval in which θ\theta lies is

A

π2≤θ≤3π4\frac{\pi}{2} \leq \theta \leq \frac{3\pi}{4}

B

0≤θ≤π40 \leq \theta \leq \frac{\pi}{4}

C

−π4≤θ≤0- \frac{\pi}{4} \leq \theta \leq 0

D

π4≤θ≤π2\frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}

Answer

0≤θ≤π40 \leq \theta \leq \frac{\pi}{4}

Explanation

Solution

θ=sin⁡−1x+cos⁡−1x−tan⁡−1x=π2−tan⁡−1x\theta = \sin^{- 1}x + \cos^{- 1}x - \tan^{- 1}x = \frac{\pi}{2} - \tan^{- 1}xWe know

−π2<tan⁡−1x<π2⇒π2>−tan⁡−1x>−π2∴0<π2−tan⁡−1x<π4- \frac{\pi}{2} < \tan^{- 1}x < \frac{\pi}{2} \Rightarrow \frac{\pi}{2} > - \tan^{- 1}x > - \frac{\pi}{2}\therefore 0 < \frac{\pi}{2} - \tan^{- 1}x < \frac{\pi}{4}