If $\alpha$ and $\beta$ are the roots of the quadratic equation $3x^2 - 16x + 5 = 0$,then $\tan^{-1} \alpha + \tan^{-1} \beta - \tan^{-1}\left(\frac{\alpha + \beta}{1 - \alpha \beta}\right) = $

Let $x = \frac{m}{n}$ ($m, n$ are co-prime natural numbers) be a solution of the equation $\cos(2 \sin^{-1} x) = \frac{1}{9}$ and let $\alpha, \beta$ $(\alpha > \beta)$ be the roots of the equation $mx^2 - nx - m + n = 0$. Then the point $(\alpha, \beta)$ lies on the line

The equation $\cos ^{-1}(1-x)-2 \cos ^{-1} x=\frac{\pi}{2}$ has

If $\tan ^{-1} \frac{1}{5}+\frac{1}{2} \sec ^{-1} x+\tan ^{-1} \frac{1}{8}=\frac{\pi}{8}$, then $x^2=$

If $f(x) = \cot^{-1} \left( \frac{3x - x^3}{1 - 3x^2} \right)$ and $g(x) = \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right)$,then for $0 < a < \frac{1}{\sqrt{3}}$,the value of $\lim_{x \to a} \frac{f(x) - f(a)}{g(x) - g(a)}$ is

If $k = \tan(\frac{\pi}{4} + \frac{1}{2}\cos^{-1}(\frac{2}{3})) + \tan(\frac{1}{2}\sin^{-1}(\frac{2}{3}))$, then the number of solutions of the equation $\sin^{-1}(kx-1) = \sin^{-1}x - \cos^{-1}x$ is . . . . . . .