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Question: If \(0 \leq x \leq 1\) and \(\theta = \sin^{- 1}x + \cos^{- 1}x - \tan^{- 1}x\), then...

If 0≤x≤10 \leq x \leq 1 and θ=sin⁡−1x+cos⁡−1x−tan⁡−1x\theta = \sin^{- 1}x + \cos^{- 1}x - \tan^{- 1}x, then

A

θ≤π/2

B

θ≥π/4

C

θ = π/4

D

π/4 ≤θ≤π/2

Answer

π/4 ≤θ≤π/2

Explanation

Solution

θ=sin⁡−1x+cos⁡−1x−tan⁡−1x\theta = \sin^{- 1}x + \cos^{- 1}x - \tan^{- 1}x

⇒ θ=π2−tan⁡−1x\theta = \frac{\pi}{2} - \tan^{- 1}xand since 0≤tan⁡−1x≤π40 \leq \tan^{- 1}x \leq \frac{\pi}{4} when

0≤x≤10 \leq x \leq 1we finedπ4≤θ≤π2\frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}