If $\cos ^{-1} x>\sin ^{-1} x$, then
The principal value of $\cos ^{-1}\left(-\frac{1}{2}\right)$ is ........... .
$$\begin{gathered} \because\quad 0 \leq \cos ^{-1} x \leq \pi \\ \cos ^{-1}\left(-\frac{1}{2}\right)=\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)=\frac{2 \pi}{3} \end{gathered}$$
The value of $\sin ^{-1}\left(\sin \frac{3 \pi}{5}\right)$ is ........ .
$$\begin{aligned} & \because\quad-\frac{\pi}{2} \leq \sin ^{-1} x \leq \frac{\pi}{2} \\ & \therefore \quad \sin ^{-1}\left(\sin \frac{3 \pi}{5}\right)=\sin ^{-1} \sin \left(\pi-\frac{2 \pi}{5}\right)=\sin ^{-1}\left(\sin \frac{2 \pi}{5}\right)=\frac{2 \pi}{5} \end{aligned} $$
If $\cos \left(\tan ^{-1} x+\cot ^{-1} \sqrt{3}\right)=0$, then the value of $x$ is ............ .
$$\begin{array}{l} \text { We have, } & \cos \left(\tan ^{-1} x+\cot ^{-1} \sqrt{3}\right)=0 \\ \Rightarrow & \tan ^{-1} x+\cot ^{-1} \sqrt{3}=\cos ^{-1} 0 \\ \Rightarrow & \tan ^{-1} x+\cot ^{-1} \sqrt{3}=\cos ^{-1} \cos \frac{\pi}{2} \\ \Rightarrow & \tan ^{-1} x+\cot ^{-1} \sqrt{3}=\frac{\pi}{2} \\ \Rightarrow & \tan ^{-1} x=\frac{\pi}{2}-\cot ^{-1} \sqrt{3} \quad \left[\because \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right]\\ \Rightarrow & \tan ^{-1} x=\tan ^{-1} \sqrt{3} \\ \therefore & x=\sqrt{3} \end{array}$$
The set of values of $\sec ^{-1} \frac{1}{2}$ is ........... .
Since, domain of $\sec ^{-1} x$ is $R-(-1,1)$.
$$\Rightarrow \quad(-\infty,-1] \cup[1, \infty)$$
So, there is no set of values exist for $\sec ^{-1} \frac{1}{2}$.
So, $\phi$ is the answer.