Question

Prove that
(i) ${\tan }^{-1}\left(\frac{1-x^2}{2x}\right)+{\cot }^{-1}\left(\frac{1-{\mathrm{x}}^2}{2\mathrm{x}}\right)=\frac{\mathrm{\pi }}{2}$

(ii) $\sin \left\{{\tan }^{-1}\frac{1-x^2}{2x}+{\cos }^{-1}\frac{1-x^2}{2x}\right\}=1$

Answer 1

(i)
$$\begin{gathered} {\tan }^{-1}\left(\frac{1-x^2}{2x}\right)+{\cot }^{-1}\left(\frac{1-x^2}{2x}\right)=\frac{\mathrm{\pi }}{2} \\ \mathrm{LHS}={\tan }^{-1}\left(\frac{1-x^2}{2x}\right)+{\cot }^{-1}\left(\frac{1-x^2}{2x}\right) \\ ={\tan }^{-1}\left(\frac{1-x^2}{2x}\right)+\frac{\mathrm{\pi }}{2}-{\tan }^{-1}\left(\frac{1-x^2}{2x}\right)\left[\because {\tan }^{-1}x+{\cot }^{-1}x=\frac{\mathrm{\pi }}{2}\right] \\ =\frac{\mathrm{\pi }}{2}=\mathrm{RHS} \end{gathered}$$

(ii)
$$\begin{gathered} \sin \left\{{\tan }^{-1}\left(\frac{1-x^2}{2x}\right)+{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right\}=1 \\ \mathrm{LHS}=\sin \left\{{\tan }^{-1}\left(\frac{1-x^2}{2x}\right)+{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right\} \\ =\sin \left\{{\sin }^{-1}\left(\frac{\frac{1-x^2}{2x}}{\sqrt{1+\frac{1-x^2}{2x}}}\right)+{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right\}\left[\because {\tan }^{-1}x={\sin }^{-1}\frac{x}{\sqrt{1+x^2}}\right] \\ =\sin \left\{{\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right)+{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right\} \\ =\sin \left\{\frac{\mathrm{\pi }}{2}\right\}\left[\because {\sin }^{-1}x+{\cos }^{-1}x=\frac{\mathrm{\pi }}{2}\right] \\ =1=\mathrm{RHS} \end{gathered}$$