Match List-I with List-II regarding the value of the following.
| List-I | List-II | ||
| (A) | $cos^{-1}(cos\frac{7\pi }{6})$ | (I) | $\frac{\pi }{2}$ |
| (B) | $cos^{-1}(cos\frac{5\pi }{4})$ | (II) | $\frac{\pi }{4}$ |
| (C) | $sin^{-1}\frac{4}{5}+2tan^{-1}\frac{1}{3}$ | (III) | $\frac{5\pi }{6}$ |
| (D) | $tan^{-1}\frac{x}{y}-tan^{-1}\frac{x-y}{x+y}$ | (IV) | $\frac{3\pi }{4}$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
(A) $\cos^{-1}(\cos\frac{7\pi }{6})=\cos^{-1}(-\frac{\sqrt{3}}{2})$
$=\pi - \frac{\pi}{6}=\frac{5\pi }{6}$ (III)
(B) $\cos^{-1}(\cos\frac{5\pi }{4})=\cos^{-1}(-\frac{1}{\sqrt{2}})$
$=\pi - \frac{\pi}{4}=\frac{3\pi }{4}$ (IV)
(C) $\sin^{-1}\frac{4}{5}+2\tan^{-1}\frac{1}{3}$
$=\tan^{-1}\frac{4}{3}+\tan^{-1}\frac{2×\frac{1}{3}}{1-\frac{1}{9}}$
$=\tan^{-1}\frac{4}{3}+\tan^{-1}\frac{3}{4}=\frac{\pi}{2}$ (I)
(D) $\tan^{-1}\frac{x}{y}-\tan^{-1}\frac{x-y}{x+y}$
$=\tan^{-1}\frac{x}{y}-\tan^{-1}\frac{\frac{x}{y}-1}{\frac{x}{y}+1}=\tan^{-1}\frac{x}{y}-\tan^{-1}\frac{x}{y}+\tan^{-1}1=\frac{\pi}{4}$ (II)