Inverse Trigonometric Function

The principal solution of $\tan^{-1}\left( \tan\left( \dfrac{7\pi}{6} \right) \right)$ is

The value of $\tan ^{2}\left(\sec ^{-1} 4\right)+\cot ^{2}\left(\operatorname{cosec}^{-1} 3\right)$ is

$\dfrac{\tan^{-1}(\sqrt{3})-{\mathrm{sec}}^{-1}(-2)}{cosec^{-1}(-\sqrt{2})+\cos^{-1}\left( -\dfrac{1}{2} \right)}$ is equal to

Find the value of $\sin\left[ \tan^{-1}(-\sqrt{3})+\cos^{-1}\left( -\dfrac{\sqrt{3}}{2} \right) \right]$.

If $\sin ^{-1}\left(\dfrac{x}{13}\right)+\operatorname{cosec}^{-1}\left(\dfrac{13}{12}\right)=\dfrac{\pi}{2}$, then the value of

The value of $\tan ^{-1}(-\sqrt{3})-\sin ^{-1}\left(\dfrac{1}{\sqrt{2}}\right)+\cos ^{-1}\left(\dfrac{-1}{2}\right)$ is

The value of $\tan ^{-1}\left(\tan \dfrac{7 \pi}{6}\right)$ is

$\tan \left(\cos ^{-1} \dfrac{1}{\sqrt{2}}+\tan ^{-1} \dfrac{1}{2}\right)=$

If $a=\sin^{-1}(\sin(5))$ and $b=\cos^{-1}(\cos(5))$, then $a^{2}+b^{2}$ is equal to

$\cos \left(\cos ^{-1}\left(-\dfrac{1}{2}\right)+\dfrac{\pi}{3}\right)=$

$\sec ^{2}\left(\tan ^{-1} 3\right)-\tan ^{2}\left(\sec ^{-1} 3\right)=$

If $\sin ^{-1}\left(\tan \dfrac{\pi}{4}\right)-\sin ^{-1}\left(\sqrt{\dfrac{3}{x}}\right)=\dfrac{\pi}{6}$ then $x$ is a root of the equation

$\cot ^{-1}\left(2 \cos \left(2 \operatorname{cosec}^{-1}(\sqrt{2})\right)\right)=\ldots$

If $\sin ^{-1}\left(\dfrac{x}{5}\right)+\operatorname{cosec}^{-1}\left(\dfrac{5}{4}\right)=\dfrac{\pi}{2}$, then the value of $x$ is