Solve the following equation for x. 
$\cos \left({\tan }^{-1}\mathrm{x}\right)=\sin \left({\cot }^{-1}\left(\frac{3}{4}\right)\right)$

$\text{ We have, }\cos \left({\tan }^{-1}\mathrm{x}\right)=\sin \left({\cot }^{-1}\frac{3}{4}\right)$
$\mathrm{Let} {\tan }^{-1}\mathrm{x}=\mathrm{\theta }\text{ and }{\cot }^{-1}\frac{3}{4}=\mathrm{\Phi }\mathrm{\forall }\mathrm{\theta }\in \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$
$\text{ and }\mathrm{\Phi }\in (0,\mathrm{\pi })$
$\begin{array}{l}\Rightarrow \tan \mathrm{\theta }=\mathrm{x}\text{ and }\cot \mathrm{\Phi }=\frac{3}{4}\\ \Rightarrow \sec \mathrm{\theta }=\sqrt{1+{\tan }^2\mathrm{\theta }}\text{ and }\mathrm{cosec}\mathrm{\Phi }=\sqrt{1+{\cot }^2\mathrm{\varphi }}\end{array}$
$\text{ [taking positive square root as }\left.\mathrm{\theta }\in \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)\text{ and }\mathrm{\Phi }\in (\mathrm{\theta },\mathrm{\pi })\right]$
$\begin{array}{l}\Rightarrow \sec \mathrm{\theta }=\sqrt{1+{\mathrm{x}}^2}\\ \text{ and }\mathrm{cosec}\mathrm{\varphi }=\sqrt{1+{\left(\frac{3}{4}\right)}^2}=\sqrt{\frac{16+9}{16}}=\sqrt{\frac{25}{16}}=\frac{5}{4}\\ \Rightarrow \frac{1}{\cos \mathrm{\theta }}=\sqrt{1+{\mathrm{x}}^2}\text{ and }\frac{1}{\sin \mathrm{\varphi }}=\frac{5}{4}\\ \Rightarrow \cos \mathrm{\theta }=\frac{1}{\sqrt{1+{\mathrm{x}}^2}}\text{ and }\sin \mathrm{\varphi }=\frac{4}{5}\\ \Rightarrow \mathrm{\theta }={\cos }^{-1}\frac{1}{\sqrt{1+{\mathrm{x}}^2}}\text{ and }\mathrm{\varphi }={\sin }^{-1}\frac{4}{5}\\ \Rightarrow {\tan }^{-1}\mathrm{x}={\cos }^{-1}\frac{1}{\sqrt{1+{\mathrm{x}}^2}}\text{ and }{\cot }^{-1}\frac{3}{4}={\sin }^{-1}\frac{4}{5}\end{array}$
On substituting these values in Eq. (i), we get
$\begin{array}{l}\cos \left({\cos }^{-1}\frac{1}{\sqrt{1+{\mathrm{x}}^2}}\right)=\sin \left({\sin }^{-1}\frac{4}{5}\right)\\ \Rightarrow \frac{1}{\sqrt{1+{\mathrm{x}}^2}}=\frac{4}{5}\left[\because \cos \left({\cos }^{-1}\mathrm{x}\right)=\mathrm{x};\mathrm{\forall }\mathrm{x}\in [-1,1]\right.\\ \left.\text{ and }\sin \left({\sin }^{-1}\mathrm{x}\right)=\mathrm{x};\mathrm{\forall }\mathrm{x}\in [-1,1]\right]\end{array}$
On squaring both sides, we get
$16\left({\mathrm{x}}^2+1\right)=25\Rightarrow 16{\mathrm{x}}^2=9\Rightarrow {\mathrm{x}}^2=\frac{9}{16}$
$\Rightarrow \mathrm{x}=\pm \frac{3}{4}$  [taking square root both sides]
But $\mathrm{x}=\frac{-3}{4}$ does not satisfy the given equation. Hence, the required solution is $\mathrm{x}=\frac{3}{4}$