Match List - I with List - II.

Choose the correct answer from the options given below :

(A) - (IV), (B) - (I), (C) - (III), (D) - (II)

(A) $tan^{-1}\sqrt{3}-sec^{-1}(-2)$

Let $tan^{-1}\sqrt{3}=x$  i.e.

$tanx=\sqrt{3}⇒tan\frac{\pi}{3}$

$∴x=\frac{\pi}{3}∈[-\frac{\pi}{2},\frac{\pi}{2}]$

Now, let $sec^{-1}(-2) = y$ that is sec y = -2

$∴y=-sec\frac{\pi}{3}$ i.e.

$secy=sec(\pi-\frac{\pi}{3})=sec\frac{2\pi}{3}$

Thus, $y = \frac{2\pi}{3}∈(0,π)-(\frac{π}{2})$

Now consider $tan^{-1}\sqrt{3}-sec^{-1}(-2)$ as:

$tan^{-1}\sqrt{3}-sec^{-1}(-2)=x-y=\frac{\pi}{3}-\frac{2\pi}{3}=\frac{-\pi}{3}$

(B) $cot^{-1}(\frac{-1}{\sqrt{3}})$

Let $y=cot^{-1}(\frac{-1}{\sqrt{3}})=coty=\frac{-1}{\sqrt{3}}$

$cot y = cot(π-\frac{π}{3})$

we know that range of principal value of $cot^{-1}x$ is (0, π)

$y=π-\frac{π}{3}⇒\frac{2π}{3}$

(C) $cos^{-1}(-\frac{1}{\sqrt{2}})$

Let $y = cos^{-1}(-\frac{1}{\sqrt{2}})$

$y =π- cos^{-1}(-\frac{1}{\sqrt{2}})$  $∴cos^{-1}(-x)=π- cos^{-1}x$

Since $cos\frac{π}{4}=\frac{1}{\sqrt{2}}$

$\frac{π}{4}=cos^{-1}(\frac{1}{\sqrt{2}})$

$y =π- cos^{-1}(\frac{1}{\sqrt{2}})$

$y=π-\frac{π}{4}⇒\frac{3π}{4}$

Range of cos^-1 is [0, π]

Principal value is $\frac{3π}{4}$

(D) $cos^{-1}(\frac{1}{2})+ sin^{-1}(\frac{1}{2})$

$cos^{-1}\frac{1}{2}+sin^{-1}(\frac{1}{2})$

$=\frac{π}{3}+\frac{π}{6}=\frac{2π+π}{6}=\frac{3π}{6}=\frac{π}{2}$

So, (A) - (II), (B) - (IV), (C) - (I), (D) - (III)

Option 4 is correct.