Inverse Trigonometric Functions

If \(\sin \left(\cot ^{-1}(1+\mathrm{x})\right)=\cos \left(\tan ^{-1} \mathrm{x}\right)\),$ then $\(\mathrm{x}=\)

Consider the following statements. I. $\sin ^{-1}\left(y^2-4 y+6\right)+\cos ^{-1}\left(y^2-4 y+6\right) =\frac{\pi}{2}, \forall y \in R$ II. $\sec ^{-1}\left(y^2-4 y+6\right)+\operatorname{cosec}^{-1}\left(y^2-4…

$\operatorname{Tanh}^{-1}(\sin \theta)=$

If $y=\operatorname{Tanh}^{-1} \sqrt{\frac{1-x}{1+x}}$, then $\frac{d y}{d x}=$

If $y=\operatorname{Tan}^{-1}\left(\frac{2 x}{1-x^2}\right),|x| < 1$, then $\left(\frac{d y}{d x}\right)_{x=\frac{1}{2}}=$

$\begin{aligned} & \text { If } \cot \left(\cos ^{-1} x\right)=\sec \left\{\tan ^{-1}\left(\frac{a}{\sqrt{b^2-a^2}}\right)\right\}: \\ & b>a, \text { then } \mathrm{x}=\end{aligned}$

The value of $x$, where $x>0$ and $\tan \left(\sec ^{-1}\left(\frac{1}{x}\right)\right)=\sin \left(\tan ^{-1} 2\right)$ is

If $y=\operatorname{Tan}^{-1} \sqrt{x^2-1}+\operatorname{Sinh}^{-1} \sqrt{x^2-1}, x>1$, then $\frac{d y}{d x}=$

If $f(x)=\operatorname{Sec}^{-1}\left(\frac{1}{2 x^2-1}\right)$ and $g(x)=\operatorname{Tan}^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$, then the derivative of $f(x)$ with respect to $g(x)$ is

If $\cos \alpha=\operatorname{sech} \beta$, then $\beta=$

The interval in which the function $f(x)=\operatorname{Tan}^{-1}(\sin x+\cos x)$ is an increasing function, is

If $\operatorname{sech}^{-1} \mathrm{x}=\log 2$ and $\operatorname{cosech}^{-1} \mathrm{y}=-\log 3$, then $(\mathrm{x}+\mathrm{y})=$