The value of $cos\left(sin^{-1}\frac{1}{4}+sec^{-1}\frac{4}{3}\right)$, is

$\frac{3\sqrt{15}-\sqrt{7}}{16}$

$cos\left(sin^{-1}\frac{1}{4}+sec^{-1}\frac{3}{4}\right)$

$=cos\left(sin^{-1}\frac{1}{4}+cos^{-1}\frac{3}{4}\right)$         $\left[∵sec^{-1}\frac{4}{3}=cos^{-1}\frac{3}{4}\right]$

$=cos\left(sin^{-1}\frac{1}{4}\right)cos\left(cos^{-1}\frac{4}{3}\right)-sin \left(sin^{-1}\frac{1}{4}\right)sin\left(cos^{-1}\frac{4}{3}\right)$

$=cos\left(cos^{-1}\frac{\sqrt{15}}{4}\right)cos\left(cos^{-1}\frac{3}{4}\right)-sin \left(sin^{-1}\frac{1}{4}\right) sin\left(sin^{-1}\frac{\sqrt{7}}{4}\right)$       $\left[∵sin^{-1}\frac{1}{4}=cos^{-1}\frac{\sqrt{15}}{4}\, and \, cos^{-1}\frac{3}{4}= sin^{-1}\frac{\sqrt{7}}{4}\right]$

$=\frac{\sqrt{15}}{4}×\frac{3}{4}-\frac{1}{4}×\frac{\sqrt{7}}{4}=\frac{3\sqrt{15}-\sqrt{7}}{16}$