Find the value of: $\;$ $\cos \left[\sin^{-1} \left(- \dfrac{4}{5}\right)\right] + \sin \left[\tan^{-1} \left(\dfrac{3}{4}\right)\right] + \cos \left[\csc^{-1} \left(\dfrac{5}{3}\right)\right]$
Let $\;$ $\sin^{-1} \left(- \dfrac{4}{5}\right) = - \sin^{-1} \left(\dfrac{4}{5}\right) = -\alpha, \hspace{1cm} -\alpha \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$
Then, $\;$ $\sin \left(\alpha\right) = \dfrac{4}{5}$ $\;\;\; \cdots \; (1)$
Let $\;$ $\tan^{-1} \left(\dfrac{3}{4}\right) = \beta, \hspace{1cm} \beta \in \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$
Then, $\;$ $\tan \left(\beta\right) = \dfrac{3}{4}$ $\;\;\; \cdots \; (2)$
Let $\;$ $\csc^{-1} \left(\dfrac{5}{3}\right) = \gamma, \hspace{1cm} \gamma \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right] - \left\{0\right\}$
Then, $\;$ $\csc \left(\gamma\right) = \dfrac{5}{3}$ $\implies$ $\sin \left(\gamma\right) = \dfrac{3}{5}$ $\;\;\; \cdots \; (3)$
From equation $(1)$, $\;$ $\cos \left(\alpha\right) = \sqrt{1 - \sin^2 \left(\alpha\right)} = \sqrt{1 - \dfrac{16}{25}} = \sqrt{\dfrac{9}{25}} = \dfrac{3}{5}$ $\;\; \cdots \; (4a)$
From equation $(2)$, $\;$ $\sin \left(\beta\right) = \dfrac{3}{\sqrt{3^2 + 4^2}} = \dfrac{3}{\sqrt{25}} = \dfrac{3}{5}$ $\;\;\; \cdots \; (4b)$
From equation $(3)$, $\;$ $\cos \left(\gamma\right) = \sqrt{1 - \sin^2\left(\gamma\right)} = \sqrt{1 - \dfrac{9}{25}} = \sqrt{\dfrac{16}{25}} = \dfrac{4}{5}$ $\;\;\; \cdots \; (4c)$
$\therefore \;$ $\cos \left[\sin^{-1} \left(- \dfrac{4}{5}\right)\right] + \sin \left[\tan^{-1} \left(\dfrac{3}{4}\right)\right] + \cos \left[\csc^{-1} \left(\dfrac{5}{3}\right)\right]$
$= \cos \left(- \alpha\right) + \sin \left(\beta\right) + \cos \left(\gamma\right)$
$= \cos \left(\alpha\right) + \sin \left(\beta\right) + \cos \left(\gamma\right)$ $\hspace{1cm} \left[\because \; \cos \left(- \theta\right) = \cos \left(\theta\right)\right]$
$= \dfrac{3}{5} + \dfrac{3}{5} + \dfrac{4}{5}$ $\hspace{1cm}$ [from equations $(4a), \; (4b), \; (4c)$]
$= \dfrac{10}{5} = 2$