Match List-I with List-II

List-I	List-II
Inverse Trigonometric function	Principal values of arguments
(A) $sin^{-1}\left(\frac{-1}{2}\right)$	(I) $\frac{-\pi}{3}$
(B) $cos^{-1}\left(\frac{-1}{2}\right)$	(II) $\frac{3\pi}{4}$
(C) $tan^{-1}(-\sqrt{3})$	(III) $\frac{-\pi}{6}$
(D) $sec^{-1}(-\sqrt{2})$	(IV) $\frac{2\pi}{3}$

Choose the correct answer from the options given below:

The principal value ranges for inverse trigonometric functions are:

$\sin^{-1}(x)$: $[-\frac{\pi}{2}, \frac{\pi}{2}]$

$\cos^{-1}(x)$: $[0, \pi]$

$\tan^{-1}(x)$: $(-\frac{\pi}{2}, \frac{\pi}{2})$

$\sec^{-1}(x)$: $[0, \pi] - \{\frac{\pi}{2}\}$

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For $\sin^{-1}(\frac{-1}{2})$:

Since $\sin(\frac{\pi}{6}) = \frac{1}{2}$, and sine is negative:

$\sin(-\frac{\pi}{6}) = -\frac{1}{2}$

The value $-\frac{\pi}{6}$ lies in the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Therefore, $\sin^{-1}(\frac{-1}{2}) = -\frac{\pi}{6}$

(A) matches with (III)

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For $\cos^{-1}(\frac{-1}{2})$:

Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$, and cosine is negative:

In the second quadrant, $\cos(\frac{2\pi}{3}) = -\frac{1}{2}$

The value $\frac{2\pi}{3}$ lies in the range $[0, \pi]$.

Therefore, $\cos^{-1}(\frac{-1}{2}) = \frac{2\pi}{3}$

(B) matches with (IV)

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For $\tan^{-1}(-\sqrt{3})$:

Since $\tan(\frac{\pi}{3}) = \sqrt{3}$, and tangent is negative:

$\tan(-\frac{\pi}{3}) = -\sqrt{3}$

The value $-\frac{\pi}{3}$ lies in the range $(-\frac{\pi}{2}, \frac{\pi}{2})$.

Therefore, $\tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}$

(C) matches with (I)

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For $\sec^{-1}(-\sqrt{2})$:

Since $\sec(\theta) = \frac{1}{\cos(\theta)}$:

$\cos(\theta) = \frac{-1}{\sqrt{2}} = \frac{-\sqrt{2}}{2}$

Since $\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$:

In the second quadrant, $\cos(\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2}$

The value $\frac{3\pi}{4}$ lies in the range $[0, \pi] - \{\frac{\pi}{2}\}$.

Therefore, $\sec^{-1}(-\sqrt{2}) = \frac{3\pi}{4}$

(D) matches with (II)

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Final matching:

(A) → (III)

(B) → (IV)

(C) → (I)

(D) → (II)