Question

Solve for the equation cos (tan^-1x) = sin(cot^-13/4)

Answer 1

$$\begin{gathered} \cos \left({\tan }^{-1}\mathrm{x}\right)=\sin \left({\cot }^{-1}\left(\frac{3}{4}\right)\right)...\left(1\right) \\ \mathrm{As}\mathrm{we}\mathrm{know}\mathrm{that}\sin \mathrm{\theta }=\cos \left(90-\mathrm{\theta }\right)\mathrm{so}, \\ \sin \left({\cot }^{-1}\frac{3}{4}\right)=\cos \left(\frac{\mathrm{\pi }}{2}-{\cot }^{-1}\left(\frac{3}{4}\right)\right) \\ \mathrm{Now}\mathrm{we}\mathrm{know}\mathrm{that}{\tan }^{-1}\mathrm{x}+{\cot }^{-1}\mathrm{x}=\frac{\mathrm{\pi }}{2}\mathrm{so} \\ \frac{\mathrm{\pi }}{2}-{\cot }^{-1}\left(\frac{3}{4}\right)={\tan }^{-1}\left(\frac{3}{4}\right)\mathrm{putting}\mathrm{this}\mathrm{we}\mathrm{get}, \\ \sin \left({\cot }^{-1}\frac{3}{4}\right)=\cos \left(\frac{\mathrm{\pi }}{2}-{\cot }^{-1}\left(\frac{3}{4}\right)\right)=\cos \left[{\tan }^{-1}\left(\frac{3}{4}\right)\right] \\ \mathrm{Now}\mathrm{putting}\mathrm{this}\mathrm{value}\mathrm{in}\mathrm{equation}\left(1\right),\mathrm{we}\mathrm{get} \\ \cos \left({\tan }^{-1}\mathrm{x}\right)=\cos \left[{\tan }^{-1}\left(\frac{3}{4}\right)\right] \\ \mathrm{or}{\tan }^{-1}\mathrm{x}={\tan }^{-1}\left(\frac{3}{4}\right) \\ \mathrm{or}\mathrm{x}=\left(\frac{3}{4}\right) \end{gathered}$$