If $\tan^{-1} (ax) + \tan^{-1} (3x) = \frac{\pi}{4}$, where $3ax^2 < 1$, then the value of $a$ for $x = \frac{1}{6}$ is...

$\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)+\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)+\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)$ is equal to

The value of $\tan ^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)$ for $|x| < \frac{1}{\sqrt{2}}, x \neq 0$ is:

If ${\sin ^{ - 1}}\left( {x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - \dots} \right) + {\cos ^{ - 1}}\left( {{x^2} - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4} - \dots} \right) = \frac{\pi }{2}$ for $0 < |x| < \sqrt 2$,then $x$ equals

$\cos \left[\cos ^{-1}\left(-\frac{1}{7}\right)+\sin ^{-1}\left(-\frac{1}{7}\right)\right]$ is equal to

The minimum value of $(\sin^{-1} x)^2 + (\cos^{-1} x)^2$ for $x \in [-1, 1]$ is: