Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| Function | Value | ||
| A. | \( \tan[\cos^{-1}(\sin(\tan^{-1}4/3))] \) | I. | \( 1/\sqrt{2} \) |
| B. | \( \sin\left(2\tan^{-1}\left(\displaystyle\frac{3}{4}\right)\right) \) | II. | 1 |
| C. | \( \cos\left[\tan^{-1}\left(\displaystyle\frac{1}{2}\right)+\tan^{-1}\left(\displaystyle\frac{1}{3}\right)\right] \) | III. | 24/25 |
| D. | \( \sin\left[\displaystyle\frac{\pi}{3}-\sin^{-1}(-1/2)\right] \) | IV. | 3/4 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-IV, B-III, C-I, D-II
(A) \( \tan[\cos^{-1}(\sin(\tan^{-1}4/3))] \)
This matches (IV) 3/4.
Let \( \tan^{-1}\frac{4}{3}=\alpha \), an acute angle in a 3-4-5 right triangle, so \( \sin\alpha=\frac{4}{5} \). Then let \( \cos^{-1}\frac{4}{5}=\beta \), again acute with sides 4, 3, 5:
\( \displaystyle \cos\beta=\frac{4}{5} \)
\( \displaystyle \sin\beta=\frac{3}{5} \)
\( \displaystyle \tan\beta=\frac{3}{4} \)
(B) \( \sin\left(2\tan^{-1}\frac{3}{4}\right) \)
This matches (III) 24/25.
Using \( \sin 2\theta=\frac{2\tan\theta}{1+\tan^2\theta} \) with \( \tan\theta=\frac{3}{4} \):
\( \displaystyle \sin 2\theta=\frac{2\cdot\frac{3}{4}}{1+\frac{9}{16}} \)
\( \displaystyle =\frac{\frac{3}{2}}{\frac{25}{16}} \)
\( \displaystyle =\frac{24}{25} \)
(C) \( \cos\left[\tan^{-1}\frac{1}{2}+\tan^{-1}\frac{1}{3}\right] \)
This matches (I) \( 1/\sqrt{2} \).
\( \displaystyle \tan^{-1}\frac{1}{2}+\tan^{-1}\frac{1}{3} \)
\( \displaystyle =\tan^{-1}\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\cdot\frac{1}{3}} \)
\( \displaystyle =\tan^{-1}\frac{5/6}{5/6} \)
\( \displaystyle =\tan^{-1}1=\frac{\pi}{4} \)
\( \displaystyle \cos\frac{\pi}{4}=\frac{1}{\sqrt{2}} \)
(D) \( \sin\left[\frac{\pi}{3}-\sin^{-1}(-1/2)\right] \)
This matches (II) 1.
\( \displaystyle \sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6} \)
\( \displaystyle \sin\left[\frac{\pi}{3}+\frac{\pi}{6}\right]=\sin\frac{\pi}{2}=1 \)
So the matching is A-IV, B-III, C-I, D-II.
Tip: The principal value of \( \sin^{-1}(-x) \) is \( -\sin^{-1}x \), so subtracting it adds the angle; forgetting this in (D) gives \( \sin\frac{\pi}{6}=\frac{1}{2} \), which is not in List-II.