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Mathematics Question on Inverse Trigonometric Functions

If 12≤x≤1,\frac{1}{2} \leq x \leq 1, then cos⁡−1x+cos⁡−1(x2+3−3x22)=\cos^{-1} x +\cos ^{-1}\left(\frac{x}{2} +\frac{\sqrt{3-3x^{2}}}{2}\right) =

A

π2 \frac{\pi}{2}

B

π4 \frac{\pi}{4}

C

π3 \frac{\pi}{3}

D

none of thesenone\, of\, these

Answer

π3 \frac{\pi}{3}

Explanation

Solution

cos⁡−1x+cos⁡−1(x2+3−3x22)\cos^{-1} x +\cos^{-1}\left(\frac{x}{2} +\frac{\sqrt{3-3x^{2}}}{2}\right)
Putting x=cos⁡θx = \cos \theta in 2nd term, we get
cos⁡−1x+cos⁡−1(cos⁡θ2+3−3cos⁡2θ2)\cos^{-1} x +\cos ^{-1} \left(\frac{\cos \theta}{2} + \frac{\sqrt{3-3\cos^{2} \theta }}{2}\right)
=cos⁡−1x+cos⁡−1(12cos⁡θ+32sin⁡θ)=\cos ^{-1} x +\cos ^{-1}\left(\frac{1}{2} \cos \theta + \frac{\sqrt{3}}{2} \sin\theta\right)
Now putting 12=rcos⁡α,32=rsin⁡α\frac{1}{2} = r \cos \alpha, \frac{\sqrt{3}}{2} = r \sin \alpha
=cos⁡−1x+cos⁡−1[rcos⁡αcos⁡θ+rsin⁡αsin⁡θ]=\cos ^{-1} x + \cos ^{-1} \left[r \cos\alpha \cos\theta +r \sin \alpha \sin \theta\right]
=cos⁡−1x+cos⁡−1[cos⁡(α−θ)][r2=14+34=44=1⇒r=1]= \cos ^{-1}x + \cos ^{-1} \left[\cos \left(\alpha -\theta\right)\right] \left[ r^{2} = \frac{1}{4} + \frac{3}{4} = \frac{4}{4} = 1 \Rightarrow r = 1\right]
=cos⁡−1+α−θ=\cos ^{-1} + \alpha -\theta
=cos⁡−1x+[−cos⁡−1x+π3]=π3= \cos ^{-1} x+ \left[- \cos ^{-1} x+ \frac{\pi}{3}\right] = \frac{\pi}{3}
[tan⁡α=rsin⁡αrcos⁡α ⇒α=π3]\begin{bmatrix}\tan\alpha = \frac{r \sin\alpha}{r \cos \alpha}\\\ \Rightarrow \alpha = \frac{\pi}{3}\end{bmatrix}