Solution

Let $\sin ^{-1}(-\frac{1}{2})=y$. Then $\sin y=-\frac{1}{2}=-\sin (\frac{\pi}{6})=\sin (-\frac{\pi}{6})$.

We know that the range of the principal value branch of $\sin ^{-1}$ is

$[-\frac{\pi}{2}, \frac{\pi}{2}]$ and $\sin (-\frac{\pi}{6})=-\frac{1}{2}$.

Therefore, the principal value of $\sin ^{-1}(-\frac{1}{2})$ is $-\frac{\pi}{6}$.

Solution

Let $\cos ^{-1}(\frac{\sqrt{3}}{2})=y$. Then, $\cos y=\frac{\sqrt{3}}{2}=\cos (\frac{\pi}{6})$.

We know that the range of the principal value branch of $\cos ^{-1}$ is

$[0, \pi]$ and $\cos (\frac{\pi}{6})=\frac{\sqrt{3}}{2}$.

Therefore, the principal value of

$ \cos ^{-1}(\frac{\sqrt{3}}{2}) \text{ is } \frac{\pi}{6} \text{. } $

Solution

Let $cosec^{-1}(2)=y$. Then,

$ cosec y=2=cosec(\frac{\pi}{6}) $

We know that the range of the principal value branch of $cosec^{-1}$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]-{0}$.

Therefore, the principal value of

$ cosec^{-1}(2) \text{ is } \frac{\pi}{6} \text{. } $

Solution

Let $\tan ^{-1}(-\sqrt{3})=y$. Then, $\tan y=-\sqrt{3}=-\tan \frac{\pi}{3}=\tan (-\frac{\pi}{3})$.

We know that the range of the principal value branch of $\tan ^{-1}$ is

$(-\frac{\pi}{2}, \frac{\pi}{2})$ and $\tan (-\frac{\pi}{3})$ is $-\sqrt{3}$.

Therefore, the principal value of

$ \tan ^{-1}(\sqrt{3}) \text{ is }-\frac{\pi}{3} \text{. } $

Solution

Let $\cos ^{-1}(-\frac{1}{2})=y$. Then, $\cos y=-\frac{1}{2}=-\cos (\frac{\pi}{3})=\cos (\pi-\frac{\pi}{3})=\cos (\frac{2 \pi}{3})$.

We know that the range of the principal value branch of $\cos ^{-1}$ is

$[0, \pi]$ and $\cos (\frac{2 \pi}{3})=-\frac{1}{2}$.

Therefore, the principal value of $\cos ^{-1}(-\frac{1}{2})$ is $\frac{2 \pi}{3}$.

Solution

Let $\tan ^{-1}(-1)=y$. Then,

$ \tan y=-1=-\tan (\frac{\pi}{4})=\tan (-\frac{\pi}{4}) . $

We know that the range of the principal value branch of $\tan ^{-1}$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$ and $\tan (-\frac{\pi}{4})=-1$.

Therefore, the principal value of $\tan ^{-1}(-1)$ is $-\frac{\pi}{4}$.

Solution

Let $\sec ^{-1}(\frac{2}{\sqrt{3}})=y$. Then, $\sec y=\frac{2}{\sqrt{3}}=\sec (\frac{\pi}{6})$.

We know that the range of the principal value branch of $\sec ^{-1}$ is

$[0, \pi]-{\frac{\pi}{2}}$ and $\sec (\frac{\pi}{6})=\frac{2}{\sqrt{3}}$.

Therefore, the principal value of $\sec ^{-1}(\frac{2}{\sqrt{3}})$ is $\frac{\pi}{6}$.

Solution

Let $\cot ^{-1}(\sqrt{3})=y$. Then, $\cot y=\sqrt{3}=\cot (\frac{\pi}{6})$.

We know that the range of the principal value branch of $\cot ^{-1}$ is $(0, \pi)$ and $\cot (\frac{\pi}{6})=\sqrt{3}$.

Therefore, the principal value of

$ \cot ^{-1}(\sqrt{3}) \text{ is } \frac{\pi}{6} \text{. } $

Solution

Let $\cos ^{-1}(-\frac{1}{\sqrt{2}})=y$. Then, $\cos y=-\frac{1}{\sqrt{2}}=-\cos (\frac{\pi}{4})=\cos (\pi-\frac{\pi}{4})=\cos (\frac{3 \pi}{4})$.

We know that the range of the principal value branch of $\cos ^{-1}$ is $[0, \pi]$ and $\cos (\frac{3 \pi}{4})=-\frac{1}{\sqrt{2}}$

Therefore, the principal value of $\cos ^{-1}(-\frac{1}{\sqrt{2}})$ is $\frac{3 \pi}{4}$.

Solution

Let $cosec^{-1}(-\sqrt{2})=y$. Then, $cosec y=-\sqrt{2}=-cosec(\frac{\pi}{4})=cosec(-\frac{\pi}{4})$.

We know that the range of the principal value branch of $cosec^{-1}$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]-{0}$ and $cosec(-\frac{\pi}{4})=-\sqrt{2}$.

Therefore, the principal value of

$ cosec^{-1}(-\sqrt{2}) \text{ is }-\frac{\pi}{4} \text{. } $

Solution

Let $\tan ^{-1}(1)=x$. Then, $\tan x=1=\tan \frac{\pi}{4}$.

$\therefore \tan ^{-1}(1)=\frac{\pi}{4}$

Let $\cos ^{-1}(-\frac{1}{2})=y$. Then, $\cos y=-\frac{1}{2}=-\cos (\frac{\pi}{3})=\cos (\pi-\frac{\pi}{3})=\cos (\frac{2 \pi}{3})$.

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

Let $\sin ^{-1}(-\frac{1}{2})=z$. Then, $\sin z=-\frac{1}{2}=-\sin (\frac{\pi}{6})=\sin (-\frac{\pi}{6})$.

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

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

$=\frac{\pi}{4}+\frac{2 \pi}{3}-\frac{\pi}{6}$

$=\frac{3 \pi+8 \pi-2 \pi}{12}=\frac{9 \pi}{12}=\frac{3 \pi}{4}$

Solution

Let $\cos ^{-1}(\frac{1}{2})=x$. Then, $\cos x=\frac{1}{2}=\cos (\frac{\pi}{3})$.

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

Let $\sin ^{-1}(\frac{1}{2})=y$. Then, $\sin y=\frac{1}{2}=\sin (\frac{\pi}{6})$.

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

$\therefore \cos ^{-1}(\frac{1}{2})+2 \sin ^{-1}(\frac{1}{2})=\frac{\pi}{3}+\frac{2 \pi}{6}=\frac{\pi}{3}+\frac{\pi}{3}=\frac{2 \pi}{3}$

Solution

It is given that $\sin ^{-1} x=y$.

We know that the range of the principal value branch of $\sin ^{-1}$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Therefore, $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$.

Solution

Let $\tan ^{-1} \sqrt{3}=x$. Then, $\tan x=\sqrt{3}=\tan \frac{\pi}{3}$.

We know that the range of the principal value branch of $\tan ^{-1}$ is $(\frac{-\pi}{2}, \frac{\pi}{2})$.

$\therefore \tan ^{-1} \sqrt{3}=\frac{\pi}{3}$

Let $\sec ^{-1}(-2)=y$. Then, $\sec y=-2=-\sec (\frac{\pi}{3})=\sec (\pi-\frac{\pi}{3})=\sec \frac{2 \pi}{3}$.

We know that the range of the principal value branch of $\sec ^{-1}$ is $[0, \pi]-{\frac{\pi}{2}}$.

$\therefore \sec ^{-1}(-2)=\frac{2 \pi}{3}$

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