$$\frac{\mathrm{d}}{\mathrm{~d} x}\left(\cos ^{-1}\left(\frac{x-\frac{1}{x}}{x+\frac{1}{x}}\right)\right)=$$

The approximate value of $\tan ^{-1}(0.999)$ is (use $\pi=3.1415$ )

The value of $\frac{\tan ^{-1}(\sqrt{3})-\sec ^{-1}(-2)}{\operatorname{cosec}^{-1}(-\sqrt{2})+\cos ^{-1}\left(\frac{-1}{2}\right)}$

If $\cot ^{-1}(7)+\cot ^{-1}(8)+\cot ^{-1}(18)=\cot ^{-1} x$, then the value of $x$ is