Question

Trigonometric Inverse Circular Function

Q. Prove that : $co{t}^{-1}\frac{1}{2}-\frac{1}{2}co{t}^{-1}\frac{4}{3}=\frac{\mathrm{\pi }}{4}$

Answer 1

$$\begin{gathered} \mathrm{Hi}, \\ {\cot }^{-1}\frac{1}{2}-\frac{1}{2}{\cot }^{-1}\frac{4}{3}=\frac{\mathrm{\pi }}{4} \\ \mathrm{It}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as} \\ 2{\cot }^{-1}\frac{1}{2}-{\cot }^{-1}\frac{4}{3}=\frac{\mathrm{\pi }}{2} \\ \mathrm{Now}\mathrm{LHS}, \\ 2{\cot }^{-1}\frac{1}{2}-{\cot }^{-1}\frac{4}{3}=2{\tan }^{-1}2-{\tan }^{-1}\frac{3}{4}\left[\mathrm{Since}{\cot }^{-1}\mathrm{x}={\tan }^{-1}\frac{1}{\mathrm{x}}\right] \\ ={\tan }^{-1}\frac{2\times 2}{1-2^2}-{\tan }^{-1}\frac{3}{4}={\tan }^{-1}\left(\frac{4}{-3}\right)-{\tan }^{-1}\frac{3}{4}\left[\mathrm{since}2{\tan }^{-1}\mathrm{x}={\tan }^{-1}\frac{2\mathrm{x}}{1-{\mathrm{x}}^2}\right] \\ =\mathrm{\pi }-{\tan }^{-1}\frac{4}{3}-{\tan }^{-1}\frac{3}{4}\left[\mathrm{since}{\tan }^{-1}\left(-\mathrm{x}\right)=\mathrm{\pi }-{\tan }^{-1}\left(\mathrm{x}\right)\right] \\ =\mathrm{\pi }-\left({\tan }^{-1}\frac{4}{3}+{\tan }^{-1}\frac{3}{4}\right) \\ =\mathrm{\pi }-{\tan }^{-1}\left(\frac{\frac{4}{3}+\frac{3}{4}}{1-\frac{4}{3}\times \frac{3}{4}}\right) \\ =\mathrm{\pi }-{\tan }^{-1}\infty =\mathrm{\pi }-\frac{\mathrm{\pi }}{2}=\frac{\mathrm{\pi }}{2}=\mathrm{RHS} \end{gathered}$$