Match List-I with List-II
|
List-I |
List-II |
|
(A) $\cos^{-1}x+\cos^{-1}(-x)$ |
(I) $\frac{\pi}{3}$ |
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(B) $cosec^{-1}(-x) + \sec^{-1}(-x)$ |
(II) $-\frac{\pi}{3}$ |
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(C) $\tan^{-1}\sqrt{3}-\sec^{-1}(-2)$ |
(III) $\pi$ |
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(D) $\tan^{-1}(\tan\frac{4π}{3})$ |
(IV) $\frac{\pi}{2}$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
|
List-I |
List-II |
|
(A) $\cos^{-1}x+\cos^{-1}(-x)$ |
(III) $\pi$ |
|
(B) $cosec^{-1}(-x) + \sec^{-1}(-x)$ |
(IV) $\frac{\pi}{2}$ |
|
(C) $\tan^{-1}\sqrt{3}-\sec^{-1}(-2)$ |
(II) $-\frac{\pi}{3}$ |
|
(D) $\tan^{-1}(\tan\frac{4π}{3})$ |
(I) $\frac{\pi}{3}$ |
(A) $\cos^{-1} x + \cos^{-1} (-x)$
Using identity:
$\cos^{-1} (-x) = \pi - \cos^{-1} x$
Therefore,
$\cos^{-1} x + \cos^{-1} (-x) = \cos^{-1} x + (\pi - \cos^{-1} x) = \pi$
Hence, (A) matches with (III) $\pi$.
(B) $\csc^{-1} (-x) + \sec^{-1} (-x)$
Recall identities:
$\csc^{-1} (-x) = -\csc^{-1} x$
$\sec^{-1} (-x) = \pi - \sec^{-1} x$ (for $x > 1$)
Adding,
$\csc^{-1} (-x) + \sec^{-1} (-x) = -\csc^{-1} x + \pi - \sec^{-1} x = \pi - (\sec^{-1} x + \csc^{-1} x)$
For particular $x$, this sum equals $\frac{\pi}{2}$.
Hence, (B) matches with (IV) $\frac{\pi}{2}$.
(C) $\tan^{-1} \sqrt{3} - \sec^{-1} (-2)$
Calculate each:
$\tan^{-1} \sqrt{3} = \frac{\pi}{3}$
$\sec^{-1} (-2) = \pi - \sec^{-1} 2 = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$
Difference:
$\frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3}$
Hence, (C) matches with (II) $-\frac{\pi}{3}$.
(D) $\tan^{-1} \left(\tan \frac{4\pi}{3}\right)$
Since $\tan^{-1}$ principal value lies in $(-\frac{\pi}{2}, \frac{\pi}{2})$, and
$\frac{4\pi}{3} = \pi + \frac{\pi}{3}$, tangent has period $\pi$ so
$\tan \frac{4\pi}{3} = \tan \frac{\pi}{3} = \sqrt{3}$
Therefore,
$\tan^{-1}(\tan \frac{4\pi}{3}) = \tan^{-1} \sqrt{3} = \frac{\pi}{3}$
Hence, (D) matches with (I) $\frac{\pi}{3}$.