Match the LIST-I with LIST-IILIST-ILIST-IIA.If cos^-1α+cos^-1β+cos^-1…

The correct answer is Option (3) → A - II, B - IV, C - III, D - I

(A) If \( \cos^{-1}\alpha+\cos^{-1}\beta+\cos^{-1}\gamma=3\pi \) then \( \alpha(2\beta+\gamma)+\beta(2\gamma+\alpha)+\gamma(2\alpha+\beta) \) equals

This matches (II) 9.

Each \( \cos^{-1} \) value is at most \( \pi \) (principal value range \( [0,\pi] \)), so a sum of \( 3\pi \) needs each to equal \( \pi \):

    \( \displaystyle \cos^{-1}\alpha=\pi\ \Rightarrow\ \alpha=-1,\ \text{and similarly}\ \beta=\gamma=-1 \)

    \( \displaystyle (-1)(-3)+(-1)(-3)+(-1)(-3)=9 \)

(B) The value of \( \tan^2(\sec^{-1}4)+\cot^2(\mathrm{cosec}^{-1}5) \) is

This matches (IV) 39.

Let \( \sec^{-1}4=\theta \) and \( \mathrm{cosec}^{-1}5=\phi \). Since \( \tan^2\theta=\sec^2\theta-1 \) and \( \cot^2\phi=\mathrm{cosec}^2\phi-1 \):

    \( \displaystyle (4^2-1)+(5^2-1)=15+24=39 \)

(C) If \( \sin\left(\cos^{-1}\frac{2}{5}+\sin^{-1}x\right)=1 \) then \( 10x \) equals

This matches (III) 4.

The sum of the two angles lies between \( -\frac{\pi}{2} \) and \( \frac{3\pi}{2} \), and in that range sine equals 1 only at \( \frac{\pi}{2} \):

    \( \displaystyle \cos^{-1}\frac{2}{5}+\sin^{-1}x=\frac{\pi}{2} \)

Since \( \sin^{-1}u+\cos^{-1}u=\frac{\pi}{2} \), this gives \( \sin^{-1}x=\sin^{-1}\frac{2}{5} \):

    \( \displaystyle x=\frac{2}{5} \)

    \( \displaystyle 10x=4 \)

(D) The length of the interval of domain of \( \sin^{-1}2x \) is

This matches (I) 1.

    \( \displaystyle -1\le 2x\le 1\ \Rightarrow\ -\frac{1}{2}\le x\le\frac{1}{2} \)

    \( \displaystyle \text{length}=\frac{1}{2}-\left(-\frac{1}{2}\right)=1 \)

So the matching is A - II, B - IV, C - III, D - I.