NTA Abhyas 2020 · Mathematics · Inverse Trigonometric Functions
D. π\piπ
∴∴∴ Let y=2tan−1(1−x22x)+2cos−1(1−x21+x2)y = 2\tan^{- 1} \left(\dfrac{1 - {x}^{2}}{2x}\right) + 2\cos^{- 1} \left(\dfrac{1 - {x}^{2}}{1 + {x}^{2}}\right)y=2tan−1(2x1−x2)+2cos−1(1+x21−x2)Putting x=tanθx = tan \thetax=tanθ
As, x∈(0,1),θ∈(0,π4)x \in \left(0,1\right), \theta \in \left(0, \dfrac{\pi}{4}\right)x∈(0,1),θ∈(0,4π) and π2−2θ∈(0,π2)\dfrac{\pi}{2} - 2\theta \in \left(0, \dfrac{\pi}{2}\right)2π−2θ∈(0,2π)∴y=2tan−1(cot2θ)+2cos−1(cos2θ)∴ y = 2\tan^{- 1} \left(\cot 2\theta\right) + 2\cos^{- 1} \left(\cos 2\theta\right)∴y=2tan−1(cot2θ)+2cos−1(cos2θ)
=2tan−1(tan(π2−2θ))+2cos−1(cos2θ)=2[π2−2θ]+4θ=π= 2\tan^{- 1} \left(\tan \left(\dfrac{\pi}{2} - 2\theta\right)\right) + 2\cos^{- 1} \left(\cos 2\theta\right) = 2\left[\dfrac{\pi}{2} - 2\theta\right] + 4\theta = \pi=2tan−1(tan(2π−2θ))+2cos−1(cos2θ)=2[2π−2θ]+4θ=π
