The value of $\tan \left(\frac{\sin ^{-1} x+\cos ^{-1} x}{2}\right)$, when $x=\frac{\sqrt{3}}{2}$, is ............... .
$\left[\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right]$
$\tan \left(\frac{\sin ^{-1} x+\cos ^{-1} x}{2}\right)=\tan \left(\frac{\pi / 2}{2}\right)$
$=\tan \frac{\pi}{4}=1$
If $y=2 \tan ^{-1} x+\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$, then $\ldots \ldots \ldots< y<$ .............. .
$$\begin{aligned} &\text { We have, }\\ &\begin{aligned} & y=2 \tan ^{-1} x+\sin ^{-1} \frac{2 x}{1+x^2} \\ \therefore\quad & y=2 \tan ^{-1} \tan \theta+\sin ^{-1} \frac{2 \tan \theta}{1+\tan ^2 \theta} \quad \quad[\text { let } x=\tan \theta] \end{aligned} \end{aligned}$$
$\Rightarrow \quad y=2 \theta+\sin ^{-1} \sin 2 \theta\quad \left[\because \sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^2 \theta}\right]$
$\Rightarrow \quad y=2 \theta+2 \theta=4 \theta \quad\left[\because \theta=\tan ^{-1} x\right]$
$$\begin{array}{ll} \Rightarrow & y=4 \tan ^{-1} x \\ \because & -\pi / 2<\tan ^{-1} x<\pi / 2 \\ \therefore & -\frac{4 \pi}{2}<4 \tan ^{-1} x<4 \pi / 2 \\ \Rightarrow & -2 \pi<4 \tan ^{-1} x<2 \pi \\ \Rightarrow & -2 \pi< y< 2 \pi \quad \left[\because y=4 \tan ^{-1} x\right] \end{array}$$
The result $\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)$ is true when the value of $x y$ is ............ .
$$\begin{aligned} \text{We know that,}\quad\tan ^{-1} x-\tan ^{-1} y & =\tan ^{-1}\left(\frac{x-y}{1+x y}\right) \\ \text{ where,}\quad x y & >-1 \end{aligned}$$
The value of $\cot ^{-1}(-x) x \in R$ in terms of $\cot ^{-1} x$ is ............ .
We know that,
$$\cot ^{-1}(-x)=\pi-\cot ^{-1} x, x \in R$$
All trigonometric functions have inverse over their respective domains.