Find the value of tan inverse(1) +cos inverse (1/4) +sin inverse (1/4)?

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$$\begin{gathered} \mathrm{S}={\tan }^{-1}\left(1\right)+{\cos }^{-1}\left(\frac{1}{4}\right)+{\sin }^{-1}\left(\frac{1}{4}\right) \\ \mathrm{As} \tan \left(\frac{\mathrm{\pi }}{4}\right)=1, \mathrm{hence} {\tan }^{-1}\left(1\right)=\frac{\mathrm{\pi }}{4} \\ \Rightarrow \mathrm{S}=\frac{\mathrm{\pi }}{4}+{\cos }^{-1}\left(\frac{1}{4}\right)+{\sin }^{-1}\left(\frac{1}{4}\right) \\ \mathrm{We} \mathrm{know} \mathrm{that} \\ {\cos }^{-1}\left(\mathrm{x}\right)+{\sin }^{-1}\left(\mathrm{x}\right)=\frac{\mathrm{\pi }}{2}, \mathrm{where} -1\le \mathrm{x}\le 1. \mathrm{hence} \\ {\cos }^{-1}\left(\frac{1}{4}\right)+{\sin }^{-1}\left(\frac{1}{4}\right)=\frac{\mathrm{\pi }}{2} \\ \Rightarrow \mathrm{S}=\frac{\mathrm{\pi }}{4}+\frac{\mathrm{\pi }}{2}=\frac{3\mathrm{\pi }}{4} \\ \mathrm{Hence}: \\ {\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\mathbf{1}\right)\mathbf{+}{\mathbf{cos}}^{\mathbf{-}\mathbf{1}}\left(\frac{\mathbf{1}}{\mathbf{4}}\right)\mathbf{+}{\mathbf{sin}}^{\mathbf{-}\mathbf{1}}\left(\frac{\mathbf{1}}{\mathbf{4}}\right)\mathbf{=}\frac{\mathbf{3}\mathbf{\pi }}{\mathbf{4}} \end{gathered}$$
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